Cube and Cube Roots

Q1: Multiple Choice Type

i. \(64\sqrt{x^6}-\sqrt{64\times x^6}\) is equal to:

Step 1: Evaluate both square roots: \[ \sqrt{x^6} = x^3,\quad \sqrt{64 \times x^6} = \sqrt{64} \cdot \sqrt{x^6} = 8x^3 \\ 64x^3 – 8x^3 = 56x^3 \] But since x is not specified, the constant part is:
Answer: d. \(56x^3\)

ii. If a number is multiplied by 3, its square will be multiplied by:

Step 1: Let the number be \(x\)
New number: \(3x\)
Original square: \(x^2\), New square: \((3x)^2 = 9x^2\)
So, square becomes 9 times
Answer: a. 9

iii. Two numbers are in the ratio 5 : 4. If the difference of their cubes is 61; the numbers are:

Step 1: Let numbers be \(5x\) and \(4x\) \[ (5x)^3 – (4x)^3 = 125x^3 – 64x^3 = 61x^3 \] Given: \(61x^3 = 61 \Rightarrow x^3 = 1 \Rightarrow x = 1\)
So, numbers = 5 and 4
Answer: a. 5 and 4

iv. The value of \(\sqrt[3]{27}+\sqrt[3]{0.008}+\sqrt[3]{0.064}\) is:

\[ \sqrt[3]{27} = 3,\quad \sqrt[3]{0.008} = 0.2,\quad \sqrt[3]{0.064} = 0.4 \\ 3 + 0.2 + 0.4 = 3.6 \] Answer: c. 3.6

v. The value of \(\sqrt[3]{\left(-3\right)^3\times8}\ \) is:

\[ (-3)^3 = -27,\quad -27 \times 8 = -216,\quad \sqrt[3]{-216} = -6 \] Answer: b. -6

vi. Statement 1: Cubes of all odd natural numbers are odd.
Statement 2: Cubes of negative integers are positive or negative integers.
Which of the following option is correct:

Step 1: Cube of any odd number (like 1, 3, 5…) is odd → TRUE
Step 2: Cube of negative integers (like \(-2^3 = -8\)) is always negative → False
Statement 1 is true, and statement 2 is false.
Answer: c. Statement 1 is true, and statement 2 is false.

vii. Assertion (A): The smallest number by which 1323 may be multiplied so that the product is a perfect cube of 7.
Reason (R): A given natural number is a perfect cube if in its prime factorization every prime occur in three times.

Step 1: Prime factorisation of 1323: \[ 1323 = 3^3 \times 7^1 \] To make all powers multiples of 3, multiply by 7 → gives \(7^3\)
⇒ \(1323 \times 7 = 9261 = (3 \times 7)^3 = 21^3\)
So A is true.
R is also a true general rule.
And R explains A correctly.
Answer: a. Both A and R are correct, and R is the correct explanation for A.

viii. Assertion (A): \(\sqrt[3]{4\frac{12}{125}}=1\frac{3}{5}\).
Reason (R): If p and q are two whole numbers (p ≠ 0), then \(\sqrt[3]{\frac{p}{q}}=\frac{\sqrt[3]{p}}{\sqrt[3]{q}}\)

Step 1: Convert to improper fraction: \[ 4\frac{12}{125} = \frac{512}{125} \Rightarrow \sqrt[3]{\frac{512}{125}} = \frac{8}{5} = 1\frac{3}{5} \] So A is true.
R is a known identity: cube root of fraction is cube root of numerator over cube root of denominator.
R is also true and explains A correctly.
Answer: a. Both A and R are correct, and R is the correct explanation for A.

ix. Assertion (A): \(\sqrt[3]{-125} = \pm25\)
Reason (R): The cube root of a negative perfect cube is negative.

Step 1: Cube root of \(-125\): \[ \sqrt[3]{-125} = -5 \Rightarrow \text{NOT } \pm25 \] So A is false.
But R is true (cube root of negative = negative number).
Answer: d. A is false, but R is true.

x. Assertion (A): \(\sqrt[3]{968} \times \sqrt[3]{1375} = 110\)
Reason (R): \(\sqrt[3]{p} \times \sqrt[3]{q} = \sqrt[3]{pq}\)

Step 1: Multiply inside the cube root: \[ \sqrt[3]{968 \times 1375} = \sqrt[3]{1331000} \] Now, is that equal to 110
So A is true.
But R is a valid property
Answer: a. Both A and R are correct, and R is the correct explanation for A.

Q2: State true or false:

i. Cube of an odd number can be even.

Step 1: Odd × Odd × Odd = Odd
So, cube of odd number is always odd, never even.
Answer: False

ii. A perfect cube does not end with two zeros.

Step 1: A number ending with 00 must have at least two 2s and two 5s as prime factors.
So, for cube to end with 00, the number must contain at least three 2s and three 5s → i.e. cube of 10 = 1000 ends with 3 zeros.
But 2 zeros? Not possible for perfect cube.
Answer: True

iii. If square of a number ends with 5, its cube will end with 25.

Step 1: Example: Number = 5, square = 25, cube = 125 (ends with 5, not 25)
Try 15: \(15^2 = 225\), \(15^3 = 3375\) → ends with 75
So cube does not always end with 25.
Answer: False

iv. The cube of a two digit number may be a three digit number.

Step 1: Try: \(10^3 = 1000\) → 4-digit
Try: \(5^3 = 125\) → 3-digit, but 5 is one-digit
Try: \(4^3 = 64\), \(6^3 = 216\), \(9^3 = 729\) → All cubes < 1000 are for one-digit numbers
So two-digit number cube ≥ 1000 (always 4-digit or more)
Answer: False

v. Cube of a natural number is called perfect cube.

Step 1: By definition, cube of any natural number = perfect cube
Ex: \(2^3 = 8\), \(3^3 = 27\), etc.
Answer: True

Q3: Find the cube roots of :

i. 110.592

Step 1: Remove decimal
110.592 = \(\frac{110592}{1000}\)
Step 2: Prime factorisation of 110592 \[ 110592 = 2^12 \times 3^3 \\ \Rightarrow \sqrt[3]{110592} = 2^4 \times 3 = 16 \times 3 = 48 \\ \sqrt[3]{\frac{110592}{1000}} = \frac{48}{10} = 4.8 \]Answer: 4.8

ii. 0.064

Step 1: Convert to fraction: \[ 0.064 = \frac{64}{1000} \\ \Rightarrow \sqrt[3]{0.064} = \sqrt[3]{\frac{64}{1000}} = \frac{\sqrt[3]{64}}{\sqrt[3]{1000}} = \frac{4}{10} \\ = 0.4 \]Answer: 0.4

Q4: Find the volume of a cubical box whose surface area is 486 cm².

Step 1: Let the side of the cube be \( a \) cm.
Surface area of a cube = \( 6a^2 \)
Given: \( 6a^2 = 486 \)
Step 2: Solve for \( a^2 \) \[ a^2 = \frac{486}{6} = 81 \\ \Rightarrow a = \sqrt{81} = 9\ \text{cm} \]Step 3: Volume of a cube = \( a^3 \) \[ a^3 = 9^3 = 729\ \text{cm}^3 \]Answer: 729 cm³

Q5: Find the cube roots of:

i. \(125 \times -64\)

Step 1: Use identity: \[ \sqrt[3]{125 \times (-64)} = \sqrt[3]{125} \times \sqrt[3]{-64} \]Step 2: Find individual cube roots: \[ \sqrt[3]{125} = 5,\quad \sqrt[3]{-64} = -4 \]Step 3: Multiply the roots: \[ 5 \times (-4) = -20 \]Answer: -20

ii. \(\frac{-125}{343}\)

Step 1: Use cube root of fraction property: \[ \sqrt[3]{\frac{-125}{343}} = \frac{\sqrt[3]{-125}}{\sqrt[3]{343}} \]Step 2: Evaluate cube roots: \[ \sqrt[3]{-125} = -5,\quad \sqrt[3]{343} = 7 \]Step 3: Final value: \[ \frac{-5}{7} \]Answer: \(-\frac{5}{7}\)

Q6: Three numbers are in the ratio 2 : 3 : 1. The sum of their cubes is 288. Find the numbers.

Step 1: Let the numbers be \(2x,\ 3x,\ 1x\)
Step 2: Write their cubes: \[ (2x)^3 = 8x^3,\quad (3x)^3 = 27x^3,\quad (x)^3 = x^3 \]Step 3: Sum of cubes: \[ 8x^3 + 27x^3 + x^3 = 36x^3 \] Given: \(36x^3 = 288\)
Step 4: Solve for \(x^3\): \[ x^3 = \frac{288}{36} = 8 \Rightarrow x = \sqrt[3]{8} = 2 \]Step 5: Find the numbers: \[ 2x = 4,\quad 3x = 6,\quad x = 2 \]Answer: The numbers are 4, 6 and 2

Q7: Find the smallest number by which 14,580 must be multiplied to make a perfect cube. Also, find the cube root of the perfect cube number obtained.

Step 1: Prime factorisation of 14,580 \[ 14580 \div 2 = 7290 \div 2 = 3645 \div 3 = 1215 \div 3 = 405 \div 3 = 135 \div 3 = 45 \div 3 = 15 \div 3 = 5 \] So, prime factors: \[ 14580 = 2^2 \times 3^6 \times 5 \]Step 2: Group the powers to form cubes
We need all powers in multiples of 3:
– \(2^2\) → needs one more 2
– \(3^6\) → already a cube
– \(5\) → needs two more 5s
So, multiply by: \(2 \times 5 \times 5 = 50\)
Step 3: Multiply to get perfect cube \[ 14580 \times 50 = 729000 \]Step 4: Cube root of perfect cube: \[ 729000 = 2^3 \times 3^6 \times 5^3 \\ \Rightarrow \sqrt[3]{729000} = 2 \times 3^2 \times 5 = 2 \times 9 \times 5 = 90 \]Answer: The smallest number is 50, and the cube root is 90

Q8: Find the smallest number by which 8,232 must be divided to make it a perfect cube. Also, find the cube root of the perfect cube so obtained.

Step 1: Prime factorisation of 8,232 \[ 8232 \div 2 = 4116 \div 2 = 2058 \div 2 = 1029 \div 3 = 343 \div 7 = 49 \div 7 = 7 \] So, \[ 8232 = 2^3 \times 3 \times 7^3 \]Step 2: Group powers to form perfect cube
– \(2^3\) → OK
– \(3\) → not complete → need to remove it (divide by 3)
– \(7^3\) → OK
So, we divide by 3 to eliminate the unpaired 3.
Step 3: New number = \( \frac{8232}{3} = 2744 \)
Step 4: Find cube root: \[ 2744 = 2^3 \times 7^3 = (2 \times 7)^3 = 14^3 \\ \Rightarrow \sqrt[3]{2744} = 14 \]Answer: The smallest number is 3, and the cube root is 14

Q9: Evaluate:

i. \( \left[\left(12^2 + 5^2\right)^{\frac{1}{2}}\right]^3 \)

Step 1: Evaluate the squares: \[ 12^2 = 144,\quad 5^2 = 25 \\ \Rightarrow 144 + 25 = 169 \]Step 2: Take square root: \[ \sqrt{169} = 13 \]Step 3: Cube the result: \[ 13^3 = 2197 \]Answer: 2197

ii. \( \left(\sqrt{10^3 – 6^3}\right)^3 \)

Step 1: Evaluate the cubes: \[ 10^3 = 1000,\quad 6^3 = 216 \\ \Rightarrow 1000 – 216 = 784 \]Step 2: Take square root: \[ \sqrt{784} = 28 \]Step 3: Cube the result: \[ 28^3 = 21952 \]Answer: 21952

Q10: 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.

Step 1: Let the cube root of the smaller number be \( x \)
Then, the smaller number = \( x^3 \)
Given: cube root of greater number = 8 ⟹ greater number = \( 8^3 = 512 \)
Step 2: Use the difference of cubes: \[ 512 – x^3 = 387 \]Step 3: Solve for \( x^3 \): \[ x^3 = 512 – 387 = 125 \Rightarrow x = \sqrt[3]{125} = 5 \]Answer: The cube root of the smaller number is 5