Selina Concise Solutions for ICSE Class 8 Mathematics Chapter 4 Cubes and Cube Roots

Exercise 4(A)

Question 1. Find the cube of:
(i) 7
(ii) 11
(iii) 16
(iv) 23
(v) 31
(vi) 42
(vii) 54
Answer:
(i) \( (7)^3 = 7 \times 7 \times 7 = 343 \)
(ii) \( (11)^3 = 11 \times 11 \times 11 = 1331 \)
(iii) \( (16)^3 = 16 \times 16 \times 16 = 4096 \)
(iv) \( (23)^3 = 23 \times 23 \times 23 = 12167 \)
(v) \( (31)^3 = 31 \times 31 \times 31 = 29791 \)
(vi) \( (42)^3 = 42 \times 42 \times 42 = 74088 \)
(vii) \( (54)^3 = 54 \times 54 \times 54 = 157464 \)
In simple words: To find the cube of any number, you multiply that number by itself three times.

Exam Tip: Memorise the cubes of numbers from 1 to 15 to solve calculations quickly and avoid errors in exams.

Question 2. Find which of the following are perfect cubes:
(i) 243
(ii) 588
(iii) 1331
(iv) 24000
(v) 1728
(vi) 1938
Answer:
(i) 243
Finding the prime factors of 243:

Here, \( 243 = 3 \times 3 \times 3 \times 3 \times 3 = (3 \times 3 \times 3) \times 3 \times 3 = 3^3 \times 3^2 \).
Since the prime factor 3 does not form a complete group of three (triplet), 243 is not a perfect cube.

(ii) 588
Finding the prime factors of 588:

Here, \( 588 = 2 \times 2 \times 7 \times 7 \times 3 \).
Since the prime factors do not form triplets, 588 is not a perfect cube.

(iii) 1331
Finding the prime factors of 1331:

Here, \( 1331 = 11 \times 11 \times 11 = 11^3 \).
Since the prime factor 11 forms a complete triplet, 1331 is a perfect cube.

(iv) 24000
Finding the prime factors of 24000:
\( 24000 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 3 \times 5 \times 5 \times 5 = (2 \times 2 \times 2) \times (2 \times 2 \times 2) \times (5 \times 5 \times 5) \times 3 = (2)^3 \times (2)^3 \times (5)^3 \times 3 \).
Since the prime factor 3 does not form a triplet, 24000 is not a perfect cube.

(v) 1728
Finding the prime factors of 1728:

Here, \( 1728 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 3 \times 3 \times 3 = (2)^3 \times (2)^3 \times (3)^3 \).
Since all prime factors can be grouped into triplets, 1728 is a perfect cube.

(vi) 1938
Finding the prime factors of 1938:

Here, \( 1938 = 2 \times 3 \times 17 \times 19 \).
Since none of the prime factors form triplets, 1938 is not a perfect cube.
In simple words: To check if a number is a perfect cube, write down its prime factors. If you can group all the factors into sets of three identical numbers with none left over, it is a perfect cube.
Exam Tip: Use prime factorisation to group factors into triplets. If any factor is left without a triplet, the number cannot be a perfect cube.

Question 3. Find the cubes of:
(i) 2.1
(ii) 0.4
(iii) 1.6
(iv) 2.5
(v) 0.12
(vi) 0.02
(vii) 0.8
Answer:
(i) \( (2.1)^3 = \left(\frac{21}{10}\right)^3 = \frac{21 \times 21 \times 21}{10 \times 10 \times 10} = \frac{9261}{1000} = 9.261 \)
(ii) \( (0.4)^3 = \left(\frac{4}{10}\right)^3 = \frac{4 \times 4 \times 4}{10 \times 10 \times 10} = \frac{64}{1000} = 0.064 \)
(iii) \( (1.6)^3 = \left(\frac{16}{10}\right)^3 = \frac{16 \times 16 \times 16}{10 \times 10 \times 10} = \frac{4096}{1000} = 4.096 \)
(iv) \( (2.5)^3 = \left(\frac{25}{10}\right)^3 = \frac{25 \times 25 \times 25}{10 \times 10 \times 10} = \frac{15625}{1000} = 15.625 \)
(v) \( (0.12)^3 = \left(\frac{12}{100}\right)^3 = \frac{12 \times 12 \times 12}{100 \times 100 \times 100} = \frac{1728}{1000000} = 0.001728 \)
(vi) \( (0.02)^3 = \left(\frac{2}{100}\right)^3 = \frac{2 \times 2 \times 2}{100 \times 100 \times 100} = \frac{8}{1000000} = 0.000008 \)
(vii) \( (0.8)^3 = \left(\frac{8}{10}\right)^3 = \frac{8 \times 8 \times 8}{10 \times 10 \times 10} = \frac{512}{1000} = 0.512 \)
In simple words: To find the cube of a decimal number, turn it into a fraction first. Then multiply both the top and bottom numbers by themselves three times, and convert it back to a decimal.

Exam Tip: Remember that if a decimal has \(d\) decimal places, its cube will have \(3 \times d\) decimal places. This is a great way to double-check your decimal point position.

Question 4. Find the cubes of:
(i) \( \frac{3}{7} \)
(ii) \( \frac{8}{9} \)
(iii) \( \frac{10}{13} \)
(iv) \( 1\frac{2}{7} \)
(v) \( 2\frac{1}{2} \)
Answer:
(i) \( \left(\frac{3}{7}\right)^3 = \frac{3 \times 3 \times 3}{7 \times 7 \times 7} = \frac{27}{343} \)
(ii) \( \left(\frac{8}{9}\right)^3 = \frac{8 \times 8 \times 8}{9 \times 9 \times 9} = \frac{512}{729} \)
(iii) \( \left(\frac{10}{13}\right)^3 = \frac{10 \times 10 \times 10}{13 \times 13 \times 13} = \frac{1000}{2197} \)
(iv) \( \left(1\frac{2}{7}\right)^3 = \left(\frac{1 \times 7 + 2}{7}\right)^3 = \left(\frac{9}{7}\right)^3 = \frac{9 \times 9 \times 9}{7 \times 7 \times 7} = \frac{729}{343} = 2\frac{43}{343} \)
(v) \( \left(2\frac{1}{2}\right)^3 = \left(\frac{2 \times 2 + 1}{2}\right)^3 = \left(\frac{5}{2}\right)^3 = \frac{5 \times 5 \times 5}{2 \times 2 \times 2} = \frac{125}{8} = 15\frac{5}{8} \)
In simple words: To find the cube of a fraction, cube the numerator and the denominator separately. For mixed numbers, convert them to improper fractions first.

Exam Tip: When dealing with mixed numbers, always change them to improper fractions first before cubing. Do not forget to convert the final answer back to a mixed fraction if needed.

Question 5. Find the cubes of:
(i) -3
(ii) -7
(iii) -12
(iv) -18
(v) -25
(vi) -30
(vii) -50
Answer:
(i) \( (-3)^3 = -3 \times -3 \times -3 = -(3 \times 3 \times 3) = -27 \)
(ii) \( (-7)^3 = -7 \times -7 \times -7 = -(7 \times 7 \times 7) = -343 \)
(iii) \( (-12)^3 = -12 \times -12 \times -12 = -(12 \times 12 \times 12) = -1728 \)
(iv) \( (-18)^3 = -18 \times -18 \times -18 = -(18 \times 18 \times 18) = -5832 \)
(v) \( (-25)^3 = -25 \times -25 \times -25 = -(25 \times 25 \times 25) = -15625 \)
(vi) \( (-30)^3 = -30 \times -30 \times -30 = -(30 \times 30 \times 30) = -27000 \)
(vii) \( (-50)^3 = -50 \times -50 \times -50 = -(50 \times 50 \times 50) = -125000 \)
In simple words: The cube of a negative number is always negative because multiplying three negative signs together gives a negative result.

Exam Tip: Remember that negative numbers raised to an odd power (like 3) always produce a negative answer, while even powers produce a positive answer.

Question 6. Which of the following are cubes of:
(i) an even number
(ii) an odd number
216, 729, 3375, 8000, 125, 343, 4096 and 9261.
Answer:
We know that the cube of an even number is always even, and the cube of an odd number is always odd.
Based on this property, we can classify the given numbers:
(i) Cubes of even numbers (even numbers among the options):
216, 8000, and 4096 are cubes of even numbers.
(ii) Cubes of odd numbers (odd numbers among the options):
729, 3375, 125, 343, and 9261 are cubes of odd numbers.
In simple words: Even numbers always have even cubes, and odd numbers always have odd cubes. So, just look at the last digit of the number to see if it is even or odd.

Exam Tip: You can quickly identify whether a cube belongs to an even or odd number by checking if the number itself is even or odd, saving time in multiple-choice questions.

Question 7. Find the least number by which 1323 must be multiplied so that the product is a perfect cube.
Answer:
First, find the prime factors of 1323:

The prime factorisation is:
\( 1323 = 3 \times 3 \times 3 \times 7 \times 7 = (3 \times 3 \times 3) \times 7 \times 7 \)
Here, the prime factor 3 forms a complete triplet, but the prime factor 7 does not. We need one more 7 to complete the triplet of 7.
Therefore, we must multiply 1323 by 7 to make it a perfect cube.
In simple words: Find the prime factors of the number. If any factor does not have a group of three, multiply by whatever is needed to complete that group. Here, we need one more 7.
Exam Tip: For multiplication questions, look for the factor that is short of completing a group of three, and multiply by that factor.

Question 8. Find the smallest number by which 8768 must be divided so that the quotient is a perfect cube.
Answer:
First, find the prime factors of 8768:

The prime factorisation is:
\( 8768 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 137 = (2 \times 2 \times 2) \times (2 \times 2 \times 2) \times 137 \)
Here, the factor 2 forms two complete triplets, but the prime factor 137 is left over without a triplet.
Therefore, we must divide 8768 by 137 to make the quotient a perfect cube.
In simple words: Write down the prime factors. Find the factor that does not have enough partners to form a group of three. Divide by that extra factor to get rid of it.
Exam Tip: For division questions, find the extra prime factors that do not form a complete group of three and divide the number by them to remove them.

Question 9. Find the smallest number by which 27783 be multiplied to get a perfect cube number.
Answer:
First, find the prime factors of 27783:

The prime factorisation is:
\( 27783 = 3 \times 3 \times 3 \times 3 \times 7 \times 7 \times 7 = (3 \times 3 \times 3) \times (7 \times 7 \times 7) \times 3 \)
Here, the prime factors 3 and 7 form one complete triplet each, but we are left with a single factor of 3. To make it a perfect cube, we need two more factors of 3 (i.e., \( 3 \times 3 \)).
Therefore, the smallest number by which 27783 must be multiplied is \( 3 \times 3 = 9 \).
In simple words: Write down the prime factors and group them into sets of three. Here, we have an extra 3. To make a complete set of three, we must multiply by two more threes, which equals 9.
Exam Tip: Always make sure to count the prime factors carefully. To make a perfect cube, find the missing factors to complete triplets and multiply by their product.

Question 10. With what least number must 8640 be divided so that the quotient is a perfect cube?
Answer:
First, find the prime factors of 8640:

The prime factorisation is:
\( 8640 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 3 \times 3 \times 3 \times 5 = (2 \times 2 \times 2) \times (2 \times 2 \times 2) \times (3 \times 3 \times 3) \times 5 \)
Here, the prime factors 2 and 3 form complete triplets, while 5 is left over without a triplet.
Therefore, we must divide 8640 by 5 to make the quotient a perfect cube.
In simple words: Write down the prime factors. Find the factor that does not have enough partners to form a group of three, and divide the number by it to get rid of it. Here, the number is 5.
Exam Tip: For division problems, locate the prime factors that are not part of any triplet, and divide by them to obtain a perfect cube.

Question 11. Which is the smallest number that must be multiplied to 77175 to make it a perfect cube?
Answer:
First, find the prime factors of 77175:

The prime factorisation is:
\( 77175 = 3 \times 3 \times 5 \times 5 \times 7 \times 7 \times 7 = (7 \times 7 \times 7) \times 3 \times 3 \times 5 \times 5 \)
Here, 7 forms a complete triplet, but we only have two 3s and two 5s. To make them complete triplets, we need one more 3 and one more 5.
Therefore, we must multiply by \( 3 \times 5 = 15 \).
In simple words: Write down the prime factors. Find the factors that do not have enough partners to form groups of three. Here, we need one more 3 and one more 5, so we multiply by 3 times 5, which is 15.
Exam Tip: If more than one prime factor lacks a complete triplet, find the missing factors for each and multiply them together to get the final multiplier.

Exercise 4(B)

Question 1. Find the cube-roots of :
(i) 64
(ii) 343
(iii) 729
(iv) 1728
(v) 9261
(vi) 4096
(vii) 8000
(viii) 3375
Answer:
We can determine the cube root of these numbers by finding their prime factors. We group the factors into sets of three identical numbers, then multiply one representative from each set.

(i) For 64:
\( 64 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \)
Grouping the prime factors into triplets:
\( 64 = (2 \times 2 \times 2) \times (2 \times 2 \times 2) \)
\( \sqrt[3]{64} = 2 \times 2 = 4 \)

(ii) For 343:
\( 343 = 7 \times 7 \times 7 \)
Grouping the prime factors into a triplet:
\( \sqrt[3]{343} = 7 \)

(iii) For 729:
\( 729 = 3 \times 3 \times 3 \times 3 \times 3 \times 3 \)
Grouping the prime factors into triplets:
\( 729 = (3 \times 3 \times 3) \times (3 \times 3 \times 3) \)
\( \sqrt[3]{729} = 3 \times 3 = 9 \)

(iv) For 1728:
\( 1728 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 3 \times 3 \times 3 \)
Grouping the prime factors into triplets:
\( 1728 = (2 \times 2 \times 2) \times (2 \times 2 \times 2) \times (3 \times 3 \times 3) \)
\( \sqrt[3]{1728} = 2 \times 2 \times 3 = 12 \)

(v) For 9261:
\( 9261 = 3 \times 3 \times 3 \times 7 \times 7 \times 7 \)
Grouping the prime factors into triplets:
\( 9261 = (3 \times 3 \times 3) \times (7 \times 7 \times 7) \)
\( \sqrt[3]{9261} = 3 \times 7 = 21 \)

(vi) For 4096:
\( 4096 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \)
Grouping the prime factors into triplets:
\( 4096 = (2 \times 2 \times 2) \times (2 \times 2 \times 2) \times (2 \times 2 \times 2) \times (2 \times 2 \times 2) \)
\( \sqrt[3]{4096} = 2 \times 2 \times 2 \times 2 = 16 \)

(vii) For 8000:
\( 8000 = 4 \times 4 \times 4 \times 5 \times 5 \times 5 \)
Grouping into triplets:
\( 8000 = (4 \times 4 \times 4) \times (5 \times 5 \times 5) \)
\( \sqrt[3]{8000} = 4 \times 5 = 20 \)

(viii) For 3375:
\( 3375 = 5 \times 5 \times 5 \times 3 \times 3 \times 3 \)
Grouping into triplets:
\( 3375 = (5 \times 5 \times 5) \times (3 \times 3 \times 3) \)
\( \sqrt[3]{3375} = 5 \times 3 = 15 \)

In simple words: To find a cube root, we break the number down into groups of three identical matching numbers. Then, we take one number from each group and multiply them together.
Exam Tip: Always make groups of three identical factors (triplets) when finding cube roots. If any factor is left without forming a complete group of three, the number is not a perfect cube.

Question 2. Find the cube-roots of :
(i) \( \frac{27}{64} \)
(ii) \( \frac{125}{216} \)
(iii) \( \frac{343}{512} \)
(iv) \( 64 \times 729 \)
(v) \( 64 \times 27 \)
(vi) \( 729 \times 8000 \)
(vii) \( 3375 \times 512 \)
Answer:
For fractions, we find the cube root of the numerator and the denominator separately. For products, we calculate the cube root of each factor individually and then multiply the outcomes.

(i) \( \sqrt[3]{\frac{27}{64}} = \frac{\sqrt[3]{27}}{\sqrt[3]{64}} = \frac{\sqrt[3]{3 \times 3 \times 3}}{\sqrt[3]{4 \times 4 \times 4}} = \frac{3}{4} \)

(ii) \( \sqrt[3]{\frac{125}{216}} = \frac{\sqrt[3]{125}}{\sqrt[3]{216}} = \frac{\sqrt[3]{5 \times 5 \times 5}}{\sqrt[3]{6 \times 6 \times 6}} = \frac{5}{6} \)

(iii) \( \sqrt[3]{\frac{343}{512}} = \frac{\sqrt[3]{343}}{\sqrt[3]{512}} = \frac{\sqrt[3]{7 \times 7 \times 7}}{\sqrt[3]{8 \times 8 \times 8}} = \frac{7}{8} \)

(iv) \( \sqrt[3]{64 \times 729} = \sqrt[3]{64} \times \sqrt[3]{729} = \sqrt[3]{4 \times 4 \times 4} \times \sqrt[3]{9 \times 9 \times 9} = 4 \times 9 = 36 \)

(v) \( \sqrt[3]{64 \times 27} = \sqrt[3]{64} \times \sqrt[3]{27} = \sqrt[3]{4 \times 4 \times 4} \times \sqrt[3]{3 \times 3 \times 3} = 4 \times 3 = 12 \)

(vi) \( \sqrt[3]{729 \times 8000} = \sqrt[3]{729} \times \sqrt[3]{8000} = \sqrt[3]{9 \times 9 \times 9} \times \sqrt[3]{20 \times 20 \times 20} = 9 \times 20 = 180 \)

(vii) \( \sqrt[3]{3375 \times 512} = \sqrt[3]{3375} \times \sqrt[3]{512} = \sqrt[3]{15 \times 15 \times 15} \times \sqrt[3]{8 \times 8 \times 8} = 15 \times 8 = 120 \)
In simple words: When finding the cube root of a fraction or a multiplication problem, you can split it up. Find the cube root of each part first, and then divide or multiply those answers.

Exam Tip: Use the property \( \sqrt[3]{\frac{a}{b}} = \frac{\sqrt[3]{a}}{\sqrt[3]{b}} \) and \( \sqrt[3]{a \times b} = \sqrt[3]{a} \times \sqrt[3]{b} \) to simplify your calculations instead of multiplying large numbers first.

Question 3. Find the cube-roots of :
(i) -216
(ii) -512
(iii) -1331
(iv) \( \frac{-27}{125} \)
(v) \( \frac{-64}{343} \)
(vi) \( \frac{512}{343} \)
(vii) -2197
(viii) -5832
(ix) -2744000
Answer:
For negative values, the cube root is always negative. We can evaluate the cube root of the positive version of the number, then add a minus sign to the final result.

(i) \( \sqrt[3]{-216} = \sqrt[3]{-6 \times -6 \times -6} = -6 \)

(ii) \( \sqrt[3]{-512} = \sqrt[3]{-8 \times -8 \times -8} = -8 \)

(iii) \( \sqrt[3]{-1331} = \sqrt[3]{-11 \times -11 \times -11} = -11 \)

(iv) \( \sqrt[3]{\frac{-27}{125}} = -\sqrt[3]{\frac{27}{125}} = -\frac{\sqrt[3]{3 \times 3 \times 3}}{\sqrt[3]{5 \times 5 \times 5}} = -\frac{3}{5} \)

(v) \( \sqrt[3]{\frac{-64}{343}} = \frac{\sqrt[3]{-64}}{\sqrt[3]{343}} = \frac{\sqrt[3]{-4 \times -4 \times -4}}{\sqrt[3]{7 \times 7 \times 7}} = -\frac{4}{7} \)

(vi) \( \sqrt[3]{\frac{512}{343}} = -\sqrt[3]{\frac{512}{343}} = -\frac{\sqrt[3]{8 \times 8 \times 8}}{\sqrt[3]{7 \times 7 \times 7}} = -\frac{8}{7} \)

(vii) For -2197:
\( \sqrt[3]{-2197} = \sqrt[3]{-13 \times -13 \times -13} = -13 \)

(viii) For -5832:
\( \sqrt[3]{-5832} = \sqrt[3]{(-2 \times -2 \times -2) \times (-3 \times -3 \times -3) \times (-3 \times -3 \times -3)} = -2 \times -3 \times -3 = -18 \)

(ix) For -2744000:
\( \sqrt[3]{-2744000} = \sqrt[3]{(-2 \times -2 \times -2) \times (-7 \times -7 \times -7) \times (-10 \times -10 \times -10)} = -2 \times -7 \times -10 = -140 \)

In simple words: The cube root of a negative number is always negative. Find the cube root of the positive number first, and then put a minus sign in front of your answer.
Exam Tip: Remember that \( \sqrt[3]{-x} = -\sqrt[3]{x} \). Don't confuse this with square roots, which cannot have negative values inside the radical in real numbers.

Question 4. Find the cube-roots of :
(i) 2.744
(ii) 9.261
(iii) 0.000027
(iv) -0.512
(v) -15.625
(vi) -125 x 1000
Answer:
To evaluate the cube root of decimals, we rewrite the decimal numbers into fractions, determine the cube roots of both the numerator and the denominator, and then convert them back into decimal notation.

(i) \( \sqrt[3]{2.744} = \sqrt[3]{\frac{2744}{1000}} = \frac{\sqrt[3]{2 \times 2 \times 2 \times 7 \times 7 \times 7}}{\sqrt[3]{10 \times 10 \times 10}} = \frac{2 \times 7}{10} = \frac{14}{10} = 1.4 \)

(ii) \( \sqrt[3]{9.261} = \sqrt[3]{\frac{9261}{1000}} = \frac{\sqrt[3]{3 \times 3 \times 3 \times 7 \times 7 \times 7}}{\sqrt[3]{10 \times 10 \times 10}} = \frac{3 \times 7}{10} = \frac{21}{10} = 2.1 \)

(iii) \( \sqrt[3]{0.000027} = \sqrt[3]{\frac{27}{1000000}} = \frac{\sqrt[3]{3 \times 3 \times 3}}{\sqrt[3]{100 \times 100 \times 100}} = \frac{3}{100} = 0.03 \)

(iv) \( \sqrt[3]{-0.512} = \sqrt[3]{\frac{-512}{1000}} = \frac{\sqrt[3]{-8 \times -8 \times -8}}{\sqrt[3]{10 \times 10 \times 10}} = \frac{-8}{10} = -0.8 \)

(v) \( \sqrt[3]{-15.625} = \sqrt[3]{\frac{-15625}{1000}} = \frac{\sqrt[3]{-(5 \times 5 \times 5) \times (5 \times 5 \times 5)}}{\sqrt[3]{10 \times 10 \times 10}} = \frac{-25}{10} = -2.5 \)

(vi) \( \sqrt[3]{-125 \times 1000} = \sqrt[3]{-(5 \times 5 \times 5) \times (10 \times 10 \times 10)} = -5 \times 10 = -50 \)
In simple words: To find the cube root of a decimal, first write it as a fraction. Find the cube root of the top and bottom numbers, then turn your final answer back into a decimal.
Exam Tip: When converting decimals to fractions, make sure the number of zeros in the denominator equals the number of decimal places in the original number (e.g., three decimal places mean dividing by 1000).

Question 5. Find the smallest number by which 26244 may be divided so that the quotient is a perfect cube.
Answer:
First, we break the number 26244 down into its prime factors:
\( 26244 = 2 \times 2 \times 3 \times 3 \times 3 \times 3 \times 3 \times 3 \times 3 \times 3 \)

Grouping the prime factors into triplets:
\( 26244 = (3 \times 3 \times 3) \times (3 \times 3 \times 3) \times 3 \times 3 \times 2 \times 2 \)

From the grouped factors, we see that \( 3 \times 3 \times 2 \times 2 \) do not form complete triplets.
To make the remaining terms a perfect cube, we divide 26244 by these extra factors:
\( 3 \times 3 \times 2 \times 2 = 36 \)

Thus, the smallest number required to divide 26244 is 36.

In simple words: Break 26244 into its prime numbers and group them in sets of three. Find the leftover numbers that did not make a full set of three, and multiply them together. That product is your answer.
Exam Tip: For division questions, the answer is the product of all factors that fail to form a group of three. Keep your prime factorization organized to avoid counting errors.

Question 6. What is the least number by which 30375 should be multiplied to get a perfect cube?
Answer:
Let us determine the prime factors of the number 30375:
\( 30375 = 3 \times 3 \times 3 \times 3 \times 3 \times 5 \times 5 \times 5 \)

Grouping these prime factors into triplets:
\( 30375 = (3 \times 3 \times 3) \times (5 \times 5 \times 5) \times 3 \times 3 \)

We observe that the prime factor 3 has an incomplete group containing only two factors (\( 3 \times 3 \)). To form a perfect triplet, we require one more 3.
Therefore, we must multiply 30375 by 3.

In simple words: Find the prime factors of 30375 and put them in groups of three. One group of three has only two numbers instead of three, so we need to multiply by one more of that number to complete the set.
Exam Tip: For multiplication questions, identify which triplet is incomplete, and find what factors are needed to complete it. The required multiplier is the product of these missing factors.

Question 7. Find the cube-roots of :
(i) 700 x 2 x 49 x 5
(ii) -216 x 1728
(iii) -64 x -125
(iv) \( \frac{-27}{343} \)
(v) \( \frac{729}{-1331} \)
(vi) 250.047
(vii) -175616
Answer:
By applying core radical properties, we analyze and evaluate each expression step-by-step:

(i) For \( 700 \times 2 \times 49 \times 5 \):
Factoring 700 gives \( 2 \times 2 \times 5 \times 5 \times 7 \). Thus,
\( 700 \times 2 \times 49 \times 5 = (2 \times 2 \times 5 \times 5 \times 7) \times 2 \times (7 \times 7) \times 5 \)
Grouping the combined factors into triplets:
\( = (2 \times 2 \times 2) \times (5 \times 5 \times 5) \times (7 \times 7 \times 7) \)
\( \sqrt[3]{700 \times 2 \times 49 \times 5} = 2 \times 5 \times 7 = 70 \)

(ii) \( \sqrt[3]{-216 \times 1728} = \sqrt[3]{-216} \times \sqrt[3]{1728} = -6 \times 12 = -72 \)

(iii) \( \sqrt[3]{-64 \times -125} = \sqrt[3]{-64} \times \sqrt[3]{-125} = -4 \times -5 = 20 \)

(iv) \( \sqrt[3]{\frac{-27}{343}} = \frac{\sqrt[3]{-27}}{\sqrt[3]{343}} = \frac{\sqrt[3]{-3 \times -3 \times -3}}{\sqrt[3]{7 \times 7 \times 7}} = -\frac{3}{7} \)

(v) \( \sqrt[3]{\frac{729}{-1331}} = \frac{\sqrt[3]{729}}{\sqrt[3]{-1331}} = \frac{\sqrt[3]{9 \times 9 \times 9}}{\sqrt[3]{-11 \times -11 \times -11}} = -\frac{9}{11} \)

(vi) For \( 250.047 \):
Converting to fraction format:
\( 250.047 = \frac{250047}{1000} \)
Prime factors of 250047 are found to be:
\( 250047 = (3 \times 3 \times 3) \times (3 \times 3 \times 3) \times (7 \times 7 \times 7) \)
\( \sqrt[3]{250.047} = \frac{\sqrt[3]{(3 \times 3 \times 3) \times (3 \times 3 \times 3) \times (7 \times 7 \times 7)}}{\sqrt[3]{10 \times 10 \times 10}} = \frac{3 \times 3 \times 7}{10} = \frac{63}{10} = 6.3 \)

(vii) For \( -175616 \):
We find the prime factorization of 175616:
\( 175616 = (2 \times 2 \times 2) \times (2 \times 2 \times 2) \times (2 \times 2 \times 2) \times (7 \times 7 \times 7) \)
\( \sqrt[3]{-175616} = -[2 \times 2 \times 2 \times 7] = -56 \)

In simple words: This section combines fractions, negative values, and products. Group the factors into triplets, take one number from each triplet, and multiply them while maintaining correct signs.
Exam Tip: Be very careful with signs! The product of two negative numbers becomes positive, but the cube root of a single negative term remains negative. Watch out for simple division errors in long prime factorization ladders.