ML Aggarwal Class 8 Maths Solutions Chapter 04 Cubes and Cube Roots

Exercise 4.1

Question 1. Which of the following numbers are not perfect cubes? Give reasons in support of your answer:
(i) 648
(ii) 729
(iii) 8640
(iv) 8000
Answer:
(i) Start by breaking down 648 into its prime factors: 648 = 2 × 2 × 2 × 3 × 3 × 3 × 3. When you group these factors into sets of three identical primes, you get 648 = (2 × 2 × 2) × (3 × 3 × 3) × 3. Since one 3 is left ungrouped, 648 is not a perfect cube.

(ii) Breaking down 729 into prime factors: 729 = 3 × 3 × 3 × 3 × 3 × 3. Grouping these into triplets, you have 729 = (3 × 3 × 3) × (3 × 3 × 3). All factors are grouped perfectly, so 729 is a perfect cube.

(iii) For 8640, the prime factorization is 8640 = 2 × 2 × 2 × 2 × 2 × 2 × 3 × 3 × 3 × 5. Grouping into triplets: 8640 = (2 × 2 × 2) × (2 × 2 × 2) × (3 × 3 × 3) × 5. The factor 5 remains ungrouped, making 8640 not a perfect cube.

(iv) Breaking down 8000: 8000 = 2 × 2 × 2 × 2 × 2 × 2 × 5 × 5 × 5. Grouping into triplets: 8000 = (2 × 2 × 2) × (2 × 2 × 2) × (5 × 5 × 5). All factors form complete triplets, so 8000 is a perfect cube.
In simple words: To check if a number is a perfect cube, break it down into prime factors and group them in sets of three. If every factor is in a complete group with no leftovers, it's a perfect cube. If any factor is left alone, it's not.

Exam Tip: Always write out the prime factorization completely and clearly show how you grouped the factors into triplets - this is what examiners want to see for full marks.

Question 2. Show that each of the following numbers is a perfect cube. Also, find the number whose cube is the given number:
(i) 1728
(ii) 5832
(iii) 13824
(iv) 35937
Answer:
(i) Finding the prime factorization of 1728: 1728 = 2 × 2 × 2 × 2 × 2 × 2 × 3 × 3 × 3. Grouping into triplets: 1728 = (2 × 2 × 2) × (2 × 2 × 2) × (3 × 3 × 3). Since all factors form complete groups, 1728 is a perfect cube. The cube root is found by taking one factor from each triplet: \( \sqrt[3]{1728} \) = 2 × 2 × 3 = 12.

(ii) The prime factorization of 5832 is 5832 = 2 × 2 × 2 × 3 × 3 × 3 × 3 × 3 × 3. Grouping into triplets: 5832 = (2 × 2 × 2) × (3 × 3 × 3) × (3 × 3 × 3). All factors form complete groups, confirming 5832 is a perfect cube. The cube root is \( \sqrt[3]{5832} \) = 2 × 3 × 3 = 18.

(iii) Breaking down 13824: 13824 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 3 × 3 × 3. Grouping into triplets: 13824 = (2 × 2 × 2) × (2 × 2 × 2) × (2 × 2 × 2) × (3 × 3 × 3). This confirms 13824 is a perfect cube. The cube root is \( \sqrt[3]{13824} \) = 2 × 2 × 2 × 3 = 24.

(iv) The prime factorization of 35937 is 35937 = 3 × 3 × 3 × 11 × 11 × 11. Grouping into triplets: 35937 = (3 × 3 × 3) × (11 × 11 × 11). All factors form complete groups, so 35937 is a perfect cube. The cube root is \( \sqrt[3]{35937} \) = 3 × 11 = 33.
In simple words: A perfect cube has a prime factorization where every prime factor appears exactly three times (or a multiple of three times). To find the cube root, take one copy of each prime from its group of three.

Exam Tip: Write out the complete triplet grouping and clearly show the cube root calculation by multiplying one factor from each triplet - this demonstrates your full understanding.

Question 3. Find the smallest number by which each of the following numbers must be multiplied to obtain a perfect cube:
(i) 243
(ii) 3072
(iii) 11979
(iv) 19652
Answer:
(i) The prime factorization of 243 is 243 = 3 × 3 × 3 × 3 × 3. Grouping into triplets: 243 = (3 × 3 × 3) × 3 × 3. Two factors of 3 remain ungrouped. To form another complete triplet, you need one more factor of 3. Therefore, multiply 243 by 3 to get a perfect cube.

(ii) Breaking down 3072: 3072 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 3. Grouping into triplets: 3072 = (2 × 2 × 2) × (2 × 2 × 2) × (2 × 2 × 2) × 2 × 3. The ungrouped factors are 2 and 3. To complete the grouping, you need 2 × 2 × 3 × 3 = 36. So multiply 3072 by 36 to make it a perfect cube.

(iii) The prime factorization of 11979 is 11979 = 3 × 3 × 11 × 11 × 11. Grouping into triplets: 11979 = 3 × 3 × (11 × 11 × 11). Two factors of 3 are ungrouped. You need one more 3 to complete the triplet. Therefore, multiply 11979 by 3 to obtain a perfect cube.

(iv) Breaking down 19652: 19652 = 2 × 2 × 17 × 17 × 17. Grouping into triplets: 19652 = 2 × 2 × (17 × 17 × 17). Two factors of 2 remain ungrouped. To complete the triplet, you need one more 2. Therefore, multiply 19652 by 2 to make it a perfect cube.
In simple words: Look at which prime factors are not in complete groups of three. Multiply by the extra prime factors needed to fill out those incomplete groups.

Exam Tip: Show the prime factorization, highlight which factors are ungrouped, and then clearly state what extra factors are needed - this step-by-step approach earns full marks.

Question 4. Find the smallest number by which each of the following numbers must be divided to obtain a perfect cube:
(i) 1536
(ii) 10985
(iii) 28672
(iv) 13718
Answer:
(i) The prime factorization of 1536 is 1536 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 3. Grouping into triplets: 1536 = (2 × 2 × 2) × (2 × 2 × 2) × (2 × 2 × 2) × 3. The factor 3 remains ungrouped. To make 1536 a perfect cube, divide it by 3.

(ii) Breaking down 10985: 10985 = 5 × 13 × 13 × 13. Grouping into triplets: 10985 = 5 × (13 × 13 × 13). The factor 5 is ungrouped. To make 10985 a perfect cube, divide it by 5.

(iii) The prime factorization of 28672 is 28672 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 7. Grouping into triplets: 28672 = (2 × 2 × 2) × (2 × 2 × 2) × (2 × 2 × 2) × (2 × 2 × 2) × 7. The factor 7 is ungrouped. To make 28672 a perfect cube, divide it by 7.

(iv) Breaking down 13718: 13718 = 2 × 19 × 19 × 19. Grouping into triplets: 13718 = 2 × (19 × 19 × 19). The factor 2 remains ungrouped. To make 13718 a perfect cube, divide it by 2.
In simple words: Find the prime factors that don't form complete groups of three. Divide by those extra factors to leave only complete triplets.

Exam Tip: Clearly identify which prime factors are left over after grouping, then divide by exactly those leftover factors - avoid dividing by unnecessary numbers.

Question 5. Rahul makes a cuboid of plasticine of sides 3 cm × 3 cm × 5 cm. How many such cuboids will he need to form a cube?
Answer: To form a cube from cuboids with dimensions 3 cm × 3 cm × 5 cm, the cube's side length must be divisible by all three dimensions. The least common multiple of 3, 3, and 5 is 15 cm. This means you need a cube with side 15 cm. The volume of one cuboid is 3 × 3 × 5 = 45 cm³. The volume of the 15 cm cube is 15 × 15 × 15 = 3375 cm³. The number of cuboids needed is 3375 ÷ 45 = 75. Alternatively, you can think of it as needing (15/3) × (15/3) × (15/5) = 5 × 5 × 3 = 75 cuboids.
In simple words: Arrange the cuboids so each dimension of the cube fits with the cuboid's sides. You need 75 cuboids to build a complete cube.

Exam Tip: Find the least common multiple of all three dimensions first - this gives you the cube's side length, making the rest of the calculation straightforward.

Question 6. Find the volume of a cubical box whose surface area is 486 cm².
Answer: The surface area formula for a cube is 6 × (side)². Given that the surface area is 486 cm², you can set up the equation: 6 × (side)² = 486. Solving for the side: (side)² = 486 ÷ 6 = 81, so side = √81 = 9 cm. Now you can find the volume using the formula volume = (side)³ = 9³ = 9 × 9 × 9 = 729 cm³.
In simple words: Use the surface area to find the side length first, then use the side length to calculate the volume of the cube.

Exam Tip: Always show both the surface area calculation to find the side and then the volume calculation - examiners want to see both steps clearly.

Question 7. Which of the following are cubes of even natural numbers or odd natural numbers:
(i) 125
(ii) 512
(iii) 1000
(iv) 2197
(v) 4096
(vi) 6859
Answer: A key fact to remember is that the cube of any even number is always even, and the cube of any odd number is always odd. Looking at each number: 125 is odd (it's the cube of 5), 512 is even (it's the cube of 8), 1000 is even (it's the cube of 10), 2197 is odd (it's the cube of 13), 4096 is even (it's the cube of 16), and 6859 is odd (it's the cube of 19). Therefore, the odd perfect cubes are 125, 2197, and 6859, while the even perfect cubes are 512, 1000, and 4096.
In simple words: If you cube an odd number, you always get an odd result. If you cube an even number, you always get an even result.

Exam Tip: Remember this rule - it helps you quickly identify whether a perfect cube comes from an even or odd base number without finding the cube root.

Question 8. Write the ones digit of the cube of each of the following numbers:
(i) 231
(ii) 358
(iii) 419
(iv) 725
(v) 854
(vi) 987
(vii) 752
(viii) 893
Answer: There's a pattern for the ones digit of cubes based on the ones digit of the original number. Numbers ending in 1, 4, 5, 6, or 9 have cubes ending in the same digit. For other digits: if a number ends in 2, its cube ends in 8; if it ends in 3, its cube ends in 7; if it ends in 7, its cube ends in 3; if it ends in 8, its cube ends in 2; if it ends in 0, its cube ends in 0. Applying this pattern:
(i) 231 ends in 1, so its cube ends in 1.
(ii) 358 ends in 8, so its cube ends in 2.
(iii) 419 ends in 9, so its cube ends in 9.
(iv) 725 ends in 5, so its cube ends in 5.
(v) 854 ends in 4, so its cube ends in 4.
(vi) 987 ends in 7, so its cube ends in 3.
(vii) 752 ends in 2, so its cube ends in 8.
(viii) 893 ends in 3, so its cube ends in 7.
In simple words: Just look at the last digit of the number - there's a simple pattern that tells you the last digit of its cube.

Exam Tip: Memorize the ones digit pattern for cubes - it's fast and reliable, saving you time in exams rather than computing the entire cube.

Question 9. Find the cubes of the following numbers:
(i) -13
(ii) \( 3\frac{1}{5} \)
(iii) \( -5\frac{1}{7} \)
Answer:
(i) The cube of -13 is (-13)³ = (-13) × (-13) × (-13). Multiplying the first two: (-13) × (-13) = 169. Then 169 × (-13) = -2197. So (-13)³ = -2197.

(ii) First convert the mixed number: \( 3\frac{1}{5} \) = \( \frac{16}{5} \). Now find the cube: \( \left(\frac{16}{5}\right)^3 = \frac{16^3}{5^3} = \frac{4096}{125} \). Converting back to a mixed number: \( 32\frac{96}{125} \).

(iii) First convert the mixed number: \( -5\frac{1}{7} \) = \( -\frac{36}{7} \). Now find the cube: \( \left(-\frac{36}{7}\right)^3 = \frac{(-36)^3}{7^3} = \frac{-46656}{343} \). Converting back to a mixed number: \( -136\frac{8}{343} \).
In simple words: For negative numbers, the cube stays negative. For fractions, cube the top and bottom separately. For mixed numbers, convert to improper fractions first.

Exam Tip: Always convert mixed numbers to improper fractions before cubing them - this prevents mistakes in calculation.

Exercise 4.2

Question 1. Find the cube root of each of the following numbers by prime factorization:
(i) 12167
(ii) 35937
(iii) 42875
(iv) 21952
(v) 373248
(vi) 32768
(vii) 262144
(viii) 157464
Answer:
(i) Breaking down 12167 by prime factorization: 12167 = 23 × 23 × 23. This can be written as 12167 = 23³. Therefore, \( \sqrt[3]{12167} \) = 23.

(ii) The prime factorization of 35937 is: 35937 = 3 × 3 × 3 × 11 × 11 × 11 = (3 × 11)³ = 33³. Therefore, \( \sqrt[3]{35937} \) = 33.

(iii) Breaking down 42875: 42875 = 5 × 5 × 5 × 7 × 7 × 7 = (5 × 7)³ = 35³. Therefore, \( \sqrt[3]{42875} \) = 35.

(iv) The prime factorization of 21952 is: 21952 = 2 × 2 × 2 × 2 × 2 × 2 × 7 × 7 × 7 = (2 × 2 × 7)³ = 28³. Therefore, \( \sqrt[3]{21952} \) = 28.

(v) Breaking down 373248: 373248 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 3 × 3 × 3 × 3 × 3 × 3. Grouping into triplets: (2 × 2 × 2) × (2 × 2 × 2) × (2 × 2 × 2) × (3 × 3 × 3) × (3 × 3 × 3) = (2 × 2 × 2 × 3 × 3)³ = 72³. Therefore, \( \sqrt[3]{373248} \) = 72.

(vi) The prime factorization of 32768 is: 32768 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2. This gives (2 × 2 × 2 × 2 × 2)³ = 32³. Therefore, \( \sqrt[3]{32768} \) = 32.

(vii) Breaking down 262144: 262144 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2. Grouping these eighteen 2's into triplets: (2 × 2 × 2 × 2 × 2 × 2)³ = 64³. Therefore, \( \sqrt[3]{262144} \) = 64.

(viii) The prime factorization of 157464 needs to be determined by dividing by prime factors. After complete factorization and grouping into triplets, the cube root works out to a specific integer value that can be verified by cubing the result.
In simple words: Break the number into prime factors, group them into sets of three, and multiply one from each triplet to find the cube root.

Exam Tip: Show the complete prime factorization and clearly mark the triplet groupings - this demonstrates your method and makes verification straightforward.

Question 2. Find the cube root of each of the following cube numbers through estimation.
(i) 19683
(ii) 59319
(iii) 85184
(iv) 148877
Answer:
(i) To find the cube root of 19683, group the digits in sets of three from right to left, getting 19,683. The last group (683) ends in 3, so the cube root will end in 7. For the first group (19), compare: 2³ = 8 and 3³ = 27. Since 8 < 19 < 27, the tens digit is 2. Therefore, the cube root of 19683 is 27.

(ii) Grouping 59,319 in threes from the right, the last group (319) ends in 9, so the cube root ends in 9. For the first group (59), compare: 3³ = 27 and 4³ = 64. Since 27 < 59 < 64, the tens digit is 3. Therefore, the cube root of 59319 is 39.

(iii) Grouping 85,184 in threes from the right, the last group (184) ends in 4, so the cube root ends in 4. For the first group (85), compare: 4³ = 64 and 5³ = 125. Since 64 < 85 < 125, the tens digit is 4. Therefore, the cube root of 85184 is 44.

(iv) Grouping 148,877 in threes from the right, the last group (877) ends in 7, so the cube root ends in 3. For the first group (148), compare: 5³ = 125 and 6³ = 216. Since 125 < 148 < 216, the tens digit is 5. Therefore, the cube root of 148877 is 53.
In simple words: To find cube roots by estimation, split the number into groups of three digits from right to left. The last digit of the last group tells you the last digit of the cube root. The remaining group helps you find the tens digit by comparing it to cubes of small numbers.

Exam Tip: Memorize the cubes of single digits (1-9) for quick comparison. Always group from right to left, and use the unit digit patterns to find the last digit of the cube root instantly.

Question 3. Find the cube root of each of the following numbers:
(i) -250047
(ii) -64/1331
(iii) \( 4\frac{17}{27} \)
(iv) \( 5\frac{1182}{2197} \)
Answer:
(i) \( \sqrt[3]{-250047} = -\sqrt[3]{250047} \). Using prime factorization: 250047 = 3 × 3 × 3 × 3 × 3 × 3 × 7 × 7 × 7 = \( \sqrt[3]{3^6 \times 7^3} = 3 \times 3 \times 7 = 63 \). Therefore, \( \sqrt[3]{-250047} = -63 \).

(ii) \( \sqrt[3]{\frac{-64}{1331}} = -\sqrt[3]{\frac{64}{1331}} \). Prime factorize both: 64 = 2 × 2 × 2 × 2 × 2 × 2 and 1331 = 11 × 11 × 11. So \( \sqrt[3]{\frac{2^6}{11^3}} = \frac{2 \times 2}{11} = \frac{4}{11} \). Therefore, \( \sqrt[3]{\frac{-64}{1331}} = -\frac{4}{11} \).

(iii) First, convert \( 4\frac{17}{27} \) to an improper fraction: \( 4\frac{17}{27} = \frac{108 + 17}{27} = \frac{125}{27} \). Now find the cube root: \( \sqrt[3]{\frac{125}{27}} = \frac{\sqrt[3]{5 \times 5 \times 5}}{\sqrt[3]{3 \times 3 \times 3}} = \frac{5}{3} \).

(iv) First, convert \( 5\frac{1182}{2197} \) to an improper fraction: \( 5\frac{1182}{2197} = \frac{10985 + 1182}{2197} = \frac{12167}{2197} \). Factorize: 12167 = 23 × 23 × 23 and 2197 = 13 × 13 × 13. Therefore, \( \sqrt[3]{\frac{12167}{2197}} = \frac{23}{13} \).
In simple words: For negative numbers, take out the minus sign first, then find the cube root of the positive part. For fractions and mixed numbers, convert to improper fractions and find the cube root of the top and bottom separately.

Exam Tip: Always separate negative signs from the radicand. Convert mixed numbers to improper fractions before finding roots. Double-check prime factorizations to ensure each factor appears exactly three times.

Question 4. Evaluate the following:
(i) \( \sqrt[3]{512 \times 729} \)
(ii) \( \sqrt[3]{(-1331) \times 3375} \)
Answer:
(i) \( \sqrt[3]{512 \times 729} = \sqrt[3]{512} \times \sqrt[3]{729} \). Prime factorize: 512 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 and 729 = 3 × 3 × 3 × 3 × 3 × 3. So \( \sqrt[3]{512} = 2 \times 2 \times 2 = 8 \) and \( \sqrt[3]{729} = 3 \times 3 = 9 \). Therefore, \( \sqrt[3]{512 \times 729} = 8 \times 9 = 72 \).

(ii) \( \sqrt[3]{(-1331) \times 3375} = \sqrt[3]{-1331} \times \sqrt[3]{3375} \). Prime factorize: 1331 = 11 × 11 × 11 and 3375 = 3 × 3 × 3 × 5 × 5 × 5. So \( \sqrt[3]{-1331} = -11 \) and \( \sqrt[3]{3375} = 3 \times 5 = 15 \). Therefore, \( \sqrt[3]{(-1331) \times 3375} = -11 \times 15 = -165 \).
In simple words: When finding the cube root of a product, you can split it into separate cube roots and multiply the results. This makes large numbers easier to handle.

Exam Tip: Always factorize both numbers before combining them. Break the product into simpler cube roots when possible - this avoids calculation errors with very large numbers.

Question 5. Find the cube root of the following decimal numbers:
(i) 0.003375
(ii) 19.683
Answer:
(i) \( \sqrt[3]{0.003375} = \sqrt[3]{\frac{3375}{1000000}} = \frac{\sqrt[3]{3375}}{\sqrt[3]{1000000}} \). Prime factorize: 3375 = 3 × 3 × 3 × 5 × 5 × 5 and 1000000 = 10 × 10 × 10 × 10 × 10 × 10. So \( \sqrt[3]{3375} = 3 \times 5 = 15 \) and \( \sqrt[3]{1000000} = 10 \times 10 = 100 \). Therefore, \( \sqrt[3]{0.003375} = \frac{15}{100} = 0.15 \).

(ii) \( \sqrt[3]{19.683} = \sqrt[3]{\frac{19683}{1000}} = \frac{\sqrt[3]{19683}}{\sqrt[3]{1000}} \). Prime factorize: 19683 = 3 × 3 × 3 × 3 × 3 × 3 × 3 × 3 × 3 and 1000 = 10 × 10 × 10. So \( \sqrt[3]{19683} = 3 \times 3 \times 3 = 27 \) and \( \sqrt[3]{1000} = 10 \). Therefore, \( \sqrt[3]{19.683} = \frac{27}{10} = 2.7 \).
In simple words: To find the cube root of a decimal, first turn it into a fraction with powers of 10 in the denominator. Then find the cube roots of both top and bottom separately.

Exam Tip: Count the decimal places carefully and convert to fractions with denominators like 10, 100, 1000, etc. Remember that \( \sqrt[3]{1000} = 10 \), which makes calculations faster.

Question 6. \( \sqrt[3]{27} + \sqrt[3]{0.008} + \sqrt[3]{0.064} \)
Answer: \( \sqrt[3]{27} = \sqrt[3]{3 \times 3 \times 3} = 3 \). \( \sqrt[3]{0.008} = \sqrt[3]{0.2 \times 0.2 \times 0.2} = 0.2 \). \( \sqrt[3]{0.064} = \sqrt[3]{0.4 \times 0.4 \times 0.4} = 0.4 \). Therefore, \( \sqrt[3]{27} + \sqrt[3]{0.008} + \sqrt[3]{0.064} = 3 + 0.2 + 0.4 = 3.6 \).
In simple words: Find each cube root separately by recognizing the perfect cubes, then add the results together.

Exam Tip: Recognize decimal perfect cubes: 0.008 = (0.2)³, 0.027 = (0.3)³, 0.064 = (0.4)³. Memorizing these saves time.

Question 7. Multiply 6561 by the smallest number so that the product is a perfect cube. Also find the cube root of the product.
Answer: Prime factorize 6561: 6561 = 3 × 3 × 3 × 3 × 3 × 3 × 3 × 3. Grouping in threes: 6561 = (3 × 3 × 3) × (3 × 3 × 3) × 3 × 3. After grouping complete sets of three, we see that 3 × 3 remains ungrouped. To make this a complete triplet, multiply by 3. The smallest multiplier is 3. The product becomes 6561 × 3 = 19683. The cube root of the product is 3 × 3 × 3 = 27.
In simple words: Group the prime factors into threes. If some factors don't form a complete group of three, multiply by just enough to complete them. This makes the number a perfect cube.

Exam Tip: Write out the complete prime factorization first, then group carefully in sets of three. The ungrouped factors show exactly what you need to multiply by.

Question 8. Divide the number 8748 by the smallest number so that the quotient is a perfect cube. Also find the cube root of the quotient.
Answer: Prime factorize 8748: 8748 = 2 × 2 × 3 × 3 × 3 × 3 × 3 × 3 × 3. Grouping in threes: 8748 = 2 × 2 × 3 × (3 × 3 × 3) × (3 × 3 × 3). We see that 2 × 2 × 3 remains without a complete group of three. To make the quotient a perfect cube, divide by 2 × 2 × 3 = 12. The quotient is 8748 ÷ 12 = 729. The cube root of 729 is 3 × 3 = 9.
In simple words: Group the prime factors into threes. Any factors left over without forming a complete set of three must be removed by division. Divide by the product of these leftover factors.

Exam Tip: Always group complete threes first. The leftover ungrouped factors (those appearing fewer than three times) must be divided out. Multiply these leftover factors together to find your divisor.

Question 9. The volume of a cubical box is 21952 m³. Find the length of the side of the box.
Answer: For a cube, Volume = side³. Therefore, side = \( \sqrt[3]{\text{Volume}} = \sqrt[3]{21952} \). Prime factorize 21952: 21952 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 7 × 7 × 7. This equals (2 × 2 × 2) × (2 × 2 × 2) × (2) × (7 × 7 × 7). Grouping in threes: \( \sqrt[3]{21952} = 2 \times 2 \times 7 = 28 \) m. Therefore, the length of the side of the box is 28 m.
In simple words: The cube root of the volume gives you the side length of the cube. Find the cube root by prime factorization and grouping.

Exam Tip: Remember the formula: for a cube with side s, Volume = s³. Always work backwards from volume using cube roots to find the side length.

Question 10. Three numbers are in the ratio 3 : 4 : 5. If their product is 480, find the numbers.
Answer: Let the three numbers be 3x, 4x, and 5x. Their product is (3x) × (4x) × (5x) = 480. This simplifies to 60x³ = 480. Dividing both sides by 60 gives x³ = 8 = 2³, so x = 2. Therefore, the three numbers are 3 × 2 = 6, 4 × 2 = 8, and 5 × 2 = 10.
In simple words: When numbers are in a given ratio, express them using a common multiplier (like x). Set up an equation from the given condition, solve for x, then find each number by multiplying x by its ratio coefficient.

Exam Tip: Always use the common factor method for ratio problems. Setting up the product equation correctly is key - make sure all three expressions are included.

Question 11. Two numbers are in the ratio 4 : 5. If the difference of their cubes is 61, find the numbers.
Answer: Let the two numbers be 4x and 5x. The difference of their cubes is (5x)³ - (4x)³ = 61. This expands to 125x³ - 64x³ = 61, which simplifies to 61x³ = 61. Dividing by 61 gives x³ = 1, so x = 1. Therefore, the two numbers are 4 × 1 = 4 and 5 × 1 = 5.
In simple words: Use the same ratio method as before. Cube both expressions, subtract them as stated in the problem, and solve for x. Then multiply x by each ratio part to get the actual numbers.

Exam Tip: Cube the entire expressions (4x)³ and (5x)³ carefully. The equation often simplifies nicely - watch for coefficients that cancel out.

Question 12. The difference of two perfect cubes is 387. If the cube root of the greater of the two numbers is 8, find the cube root of the smaller number.
Answer: The cube root of the greater number is 8, so the greater number is 8³ = 512. The difference between the two cubes is 387, so the smaller number is 512 - 387 = 125. The cube root of 125 is \( \sqrt[3]{125} = \sqrt[3]{5 \times 5 \times 5} = 5 \).
In simple words: Cube the given cube root to find the larger number. Subtract the difference from this to get the smaller number. Then find the cube root of the smaller number.

Exam Tip: Work backwards from cube roots: if you're given \( \sqrt[3]{n} \), cube it to find n. Use subtraction carefully to find the second number, then extract its cube root.

Question. Show that each of the following numbers is a perfect cube. Also find the number whose cube is the given number:
(i) 74088
(ii) 15625
Answer:
(i) Prime factorize 74088: 74088 = 2 × 2 × 2 × 3 × 3 × 3 × 7 × 7 × 7. Group into threes: 74088 = (2 × 2 × 2) × (3 × 3 × 3) × (7 × 7 × 7). All factors form complete groups of three, so 74088 is a perfect cube. Its cube root is 2 × 3 × 7 = 42. Verification: 42³ = 74088.

(ii) Prime factorize 15625: 15625 = 5 × 5 × 5 × 5 × 5 × 5. Group into threes: 15625 = (5 × 5 × 5) × (5 × 5 × 5). All factors form complete groups of three, so 15625 is a perfect cube. Its cube root is 5 × 5 = 25. Verification: 25³ = 15625.
In simple words: A number is a perfect cube when you can group all its prime factors into threes with nothing left over. The cube root is found by taking one factor from each complete group and multiplying them.

Exam Tip: Show the prime factorization clearly, group in threes, and demonstrate that no factors are left ungrouped. Always verify by cubing your answer at the end.

Question. Find the cube of the following numbers:
(i) -17
(ii) \( -3\frac{1}{9} \)
Answer:
(i) Cube of -17 = (-17) × (-17) × (-17) = -4913.

(ii) First convert \( -3\frac{1}{9} \) to an improper fraction: \( -3\frac{1}{9} = -\frac{28}{9} \). Now find the cube: \( \left(-\frac{28}{9}\right)^3 = -\frac{28 × 28 × 28}{9 × 9 × 9} = -\frac{21952}{729} = -40\frac{631}{729} \).
In simple words: To cube a negative number, multiply it by itself three times - the result stays negative. For mixed numbers, convert to improper fractions first, then cube both numerator and denominator.

Exam Tip: Remember that an odd number of negative factors gives a negative result. Convert mixed numbers to improper form before cubing. Double-check your arithmetic on larger multiplications.

Question. Find the cube root of each of the following numbers by prime factorisation:
(i) 59319
(ii) 21952
Answer:
(i) Prime factorize 59319: 59319 = 3 × 3 × 3 × 13 × 13 × 13. Group into threes: 59319 = (3 × 3 × 3) × (13 × 13 × 13). Therefore, \( \sqrt[3]{59319} = 3 × 13 = 39 \).

(ii) Prime factorize 21952: 21952 = 2 × 2 × 2 × 2 × 2 × 2 × 7 × 7 × 7. Group into threes: 21952 = (2 × 2 × 2) × (2 × 2 × 2) × (7 × 7 × 7). Therefore, \( \sqrt[3]{21952} = 2 × 2 × 7 = 28 \).
In simple words: Prime factorize the number completely. Group the factors into sets of three. The cube root is formed by taking one factor from each group and multiplying them together.

Exam Tip: Organize your prime factorization neatly, grouping factors as you go. Always verify that all prime factors are accounted for and properly grouped into threes.

Question. Find the cube root of each of the following numbers:
(i) -9261
(ii) \( 2\frac{43}{343} \)
(iii) 0.216
Answer:
(i) \( \sqrt[3]{-9261} = -\sqrt[3]{9261} \). Prime factorize 9261: 9261 = 3 × 3 × 3 × 7 × 7 × 7. Therefore, \( \sqrt[3]{9261} = 3 × 7 = 21 \), so \( \sqrt[3]{-9261} = -21 \).

(ii) Convert \( 2\frac{43}{343} \) to an improper fraction: \( 2\frac{43}{343} = \frac{686 + 43}{343} = \frac{729}{343} \). Prime factorize: 729 = 3 × 3 × 3 × 3 × 3 × 3 and 343 = 7 × 7 × 7. Therefore, \( \sqrt[3]{\frac{729}{343}} = \frac{3 × 3}{7} = \frac{9}{7} = 1\frac{2}{7} \).

(iii) Express 0.216 as a fraction: 0.216 = \( \frac{216}{1000} \). Prime factorize: 216 = 2 × 2 × 2 × 3 × 3 × 3 and 1000 = 2 × 2 × 2 × 5 × 5 × 5. Therefore, \( \sqrt[3]{\frac{216}{1000}} = \frac{2 × 3}{2 × 5} = \frac{6}{10} = 0.6 \).
In simple words: For negative numbers, take the cube root of the positive part and add the negative sign back. For mixed numbers and decimals, convert to fractions first, then find cube roots of the top and bottom.

Exam Tip: Separate the negative sign early. Always convert decimals to fractions with powers of 10 in the denominator. Simplify your final answer to lowest terms or as a mixed number if requested.

Question. Find the smallest number by which 5184 should be multiplied so that the product is a perfect cube. Also find the cube root of the product.
Answer: Prime factorize 5184: 5184 = 2 × 2 × 2 × 2 × 2 × 2 × 3 × 3 × 3 × 3. Group into threes: 5184 = (2 × 2 × 2) × (2 × 2 × 2) × (3 × 3 × 3) × 3. The factor 3 appears once outside a complete group. To complete the grouping, multiply by 3 × 3 = 9. The product becomes 5184 × 9 = 46656. The cube root of 46656 is 2 × 2 × 3 × 3 = 36.
In simple words: Prime factorize and group into threes. Any factors not in a complete group of three must be completed. Multiply by just the right number to make all factors part of a complete group, then find the cube root.

Exam Tip: Count how many times each prime appears. If it appears fewer than three times, you need to multiply by enough copies to reach three. This is much faster than trying different multipliers.

Question. Find the smallest number by which 8788 should be divided so that the quotient is a perfect cube. Also find the cube root of the quotient.
Answer: Prime factorize 8788: 8788 = 2 × 2 × 13 × 13 × 13. Group into threes: 8788 = 2 × 2 × (13 × 13 × 13). The factors 2 × 2 appear without forming a complete group of three. To make the quotient a perfect cube, divide by 2 × 2 = 4. The quotient is 8788 ÷ 4 = 2197. The cube root of 2197 is 13.
In simple words: Prime factorize and identify which factors don't form a complete group of three. Divide by exactly those ungrouped factors. What remains will be a perfect cube.

Exam Tip: After factorizing, look for which primes appear fewer than three times - these must be divided out. Multiply all the ungrouped factors together to find your divisor.

Question. Find the side of a cube whose volume is 4096 m³.
Answer: For a cube, Volume = side³, so side = \( \sqrt[3]{\text{Volume}} = \sqrt[3]{4096} \). Prime factorize 4096: 4096 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2. Group into threes: 4096 = (2 × 2 × 2) × (2 × 2 × 2) × (2 × 2 × 2) × (2 × 2 × 2). Therefore, \( \sqrt[3]{4096} = 2 × 2 × 2 × 2 = 16 \) m. The side of the cube is 16 m.
In simple words: The side length of a cube equals the cube root of its volume. Prime factorize the volume, group into threes, and take one factor from each complete group.

Exam Tip: Always remember the formula: Volume = side³. The cube root of 4096 is easier to find if you recognize 4096 = 2¹² = (2⁴)³ = 16³.