Squares and Square Roots, Cubes and Cube Roots — Exercise 3(E)

Find the cube of:

16

Answer

The cube of a number is that number raised to the power 3.

⇒ (16)³

⇒ 16 × 16 × 16

⇒ 4096

Hence, (16)³ = 4096.

Find the cube of:

30

Answer

The cube of a number is that number raised to the power 3.

⇒ (30)³

⇒ 30 × 30 × 30

⇒ 27000

Hence, (30)³ = 27000.

Find the cube of:

1.2

Answer

The cube of a number is that number raised to the power 3.

⇒ (1.2)³

⇒ 1.2 × 1.2 × 1.2

⇒ 1.728

Hence, (1.2)³ = 1.728.

Find the cube of:

0.7

Answer

The cube of a number is that number raised to the power 3.

⇒ (0.7)³

⇒ 0.7 × 0.7 × 0.7

⇒ 0.343

Hence, (0.7)³ = 0.343.

Find the cube of:

0.06

Answer

The cube of a number is that number raised to the power 3.

⇒ (0.06)³

⇒ 0.06 × 0.06 × 0.06

⇒ 0.000216

Hence, (0.06)³ = 0.000216.

Find the cube of:

$\dfrac{2}{5}$

Answer

The cube of a number is that number raised to the power 3.

$\Rightarrow \left(\dfrac{2}{5}\right)^3\\[1em] \Rightarrow \dfrac{2}{5} \times \dfrac{2}{5} \times \dfrac{2}{5}\\[1em] \Rightarrow \dfrac{8}{125}$

Hence, $\mathbf{\left(\dfrac{2}{5}\right)^3 = \dfrac{8}{125}}$.

Find the cube of:

$\dfrac{1}{9}$

Answer

The cube of a number is that number raised to the power 3.

$\Rightarrow \left(\dfrac{1}{9}\right)^3\\[1em] \Rightarrow \dfrac{1}{9} \times \dfrac{1}{9} \times \dfrac{1}{9}\\[1em] \Rightarrow \dfrac{1}{729}$

Hence, $\mathbf{\left(\dfrac{1}{9}\right)^3 = \dfrac{1}{729}}$.

Find the cube of:

$1\dfrac{3}{5}$

Answer

Converting mixed fraction into improper fraction:

$1\dfrac{3}{5} = \dfrac{8}{5}$

The cube of a number is that number raised to the power 3.

$\Rightarrow \left(1\dfrac{3}{5}\right)^3\\[1em] \Rightarrow \left(\dfrac{8}{5}\right)^3\\[1em] \Rightarrow \dfrac{8}{5} \times \dfrac{8}{5} \times \dfrac{8}{5}\\[1em] \Rightarrow \dfrac{512}{125}\\[1em] \Rightarrow 4\dfrac{12}{125}$

Hence, $\mathbf{\left(1\dfrac{3}{5}\right)^3 = 4\dfrac{12}{125}}$.

Examine whether 729 is a perfect cube. In case of perfect cube, find the cube root.

Answer

Resolving 729 into prime factors, we get :

$\begin{array}{l|r} 3 & 729 \\ \hline 3 & 243 \\ \hline 3 & 81 \\ \hline 3 & 27 \\ \hline 3 & 9 \\ \hline 3 & 3 \\ \hline & 1 \\ \end{array}$

729 = (3 × 3 × 3) × (3 × 3 × 3)

Since 729 can be expressed as the product of triplets of equal factors, so 729 is a perfect cube.

$\Rightarrow \sqrt[3]{729} = \sqrt[3]{(3 × 3 × 3) × (3 × 3 × 3)} \\[1em] \Rightarrow \sqrt[3]{729} = 3 \times 3\\[1em] \Rightarrow \sqrt[3]{729} = 9$

Hence, 729 is a perfect cube and $\mathbf{\sqrt[3]{729} = 9}$.

Examine whether 1331 is a perfect cube. In case of perfect cube, find the cube root.

Answer

Resolving 1331 into prime factors, we get :

$\begin{array}{l|r} 11 & 1331 \\ \hline 11 & 121 \\ \hline 11 & 11 \\ \hline & 1 \\ \end{array}$

1331 = (11 × 11 × 11)

Since 1331 can be expressed as the product of triplets of equal factors, so 1331 is a perfect cube.

$\Rightarrow \sqrt[3]{1331} = \sqrt[3]{11 \times 11 \times 11} \\[1em] \Rightarrow \sqrt[3]{1331} = 11$

Hence, 1331 is a perfect cube and $\mathbf{\sqrt[3]{1331} = 11}$.

Examine whether 5324 is a perfect cube. In case of perfect cube, find the cube root.

Answer

Resolving 5324 into prime factors, we get :

$\begin{array}{l|r} 2 & 5324 \\ \hline 2 & 2662 \\ \hline 11 & 1331 \\ \hline 11 & 121 \\ \hline 11 & 11 \\ \hline & 1 \\ \end{array}$

5324 = (11 × 11 × 11) × 2 × 2

Since 2 × 2 is left after grouping the prime factors in triplets, so 5324 cannot be expressed as the product of triplets of equal factors.

Hence, 5324 is not a perfect cube.

Examine whether 3375 is a perfect cube. In case of perfect cube, find the cube root.

Answer

Resolving 3375 into prime factors, we get :

$\begin{array}{l|r} 3 & 3375 \\ \hline 3 & 1125 \\ \hline 3 & 375 \\ \hline 5 & 125 \\ \hline 5 & 25 \\ \hline 5 & 5 \\ \hline & 1 \\ \end{array}$

3375 = (3 × 3 × 3) × (5 × 5 × 5)

Since 3375 can be expressed as the product of triplets of equal factors, so 3375 is a perfect cube.

$\Rightarrow \sqrt[3]{3375} = \sqrt[3]{(3 \times 3 \times 3) \times (5 \times 5 \times 5)} \\[1em] \Rightarrow \sqrt[3]{3375} = 3 \times 5\\[1em] \Rightarrow \sqrt[3]{3375} = 15$

Hence, 3375 is a perfect cube and $\mathbf{\sqrt[3]{3375} = 15}$.

Examine whether 9261 is a perfect cube. In case of perfect cube, find the cube root.

Answer

Resolving 9261 into prime factors, we get :

$\begin{array}{l|r} 3 & 9261 \\ \hline 3 & 3087 \\ \hline 3 & 1029 \\ \hline 7 & 343 \\ \hline 7 & 49 \\ \hline 7 & 7 \\ \hline & 1 \\ \end{array}$

9261 = (3 × 3 × 3) × (7 × 7 × 7)

Since 9261 can be expressed as the product of triplets of equal factors, so 9261 is a perfect cube.

$\Rightarrow \sqrt[3]{9261} = \sqrt[3]{(3 \times 3 \times 3) \times (7 \times 7 \times 7)} \\[1em] \Rightarrow \sqrt[3]{9261} = 3 \times 7\\[1em] \Rightarrow \sqrt[3]{9261} = 21$

Hence, 9261 is a perfect cube and $\mathbf{\sqrt[3]{9261} = 21}$.

Examine whether 5832 is a perfect cube. In case of perfect cube, find the cube root.

Answer

Resolving 5832 into prime factors, we get :

$\begin{array}{l|r} 2 & 5832 \\ \hline 2 & 2916 \\ \hline 2 & 1458 \\ \hline 3 & 729 \\ \hline 3 & 243 \\ \hline 3 & 81 \\ \hline 3 & 27 \\ \hline 3 & 9 \\ \hline 3 & 3 \\ \hline & 1 \\ \end{array}$

5832 = (2 × 2 × 2) × (3 × 3 × 3) × (3 × 3 × 3)

Since 5832 can be expressed as the product of triplets of equal factors, so 5832 is a perfect cube.

$\Rightarrow \sqrt[3]{5832} = \sqrt[3]{(2 \times 2 \times 2) \times (3 \times 3 \times 3) \times (3 \times 3 \times 3)} \\[1em] \Rightarrow \sqrt[3]{5832} = 2 \times 3 \times 3\\[1em] \Rightarrow \sqrt[3]{5832} = 18$

Hence, 5832 is a perfect cube and $\mathbf{\sqrt[3]{5832} = 18}$.

Examine whether 1728 is a perfect cube. In case of perfect cube, find the cube root.

Answer

Resolving 1728 into prime factors, we get :

$\begin{array}{l|r} 2 & 1728 \\ \hline 2 & 864 \\ \hline 2 & 432 \\ \hline 2 & 216 \\ \hline 2 & 108 \\ \hline 2 & 54 \\ \hline 3 & 27 \\ \hline 3 & 9 \\ \hline 3 & 3 \\ \hline & 1 \\ \end{array}$

1728 = (2 × 2 × 2) × (2 × 2 × 2) × (3 × 3 × 3)

Since 1728 can be expressed as the product of triplets of equal factors, so 1728 is a perfect cube.

$\Rightarrow \sqrt[3]{1728} = \sqrt[3]{(2 \times 2 \times 2) \times (2 \times 2 \times 2) \times (3 \times 3 \times 3)} \\[1em] \Rightarrow \sqrt[3]{1728} = 2 \times 2 \times 3\\[1em] \Rightarrow \sqrt[3]{1728} = 12$

Hence, 1728 is a perfect cube and $\mathbf{\sqrt[3]{1728} = 12}$.

Examine whether 10584 is a perfect cube. In case of perfect cube, find the cube root.

Answer

Resolving 10584 into prime factors, we get :

$\begin{array}{l|r} 2 & 10584 \\ \hline 2 & 5292 \\ \hline 2 & 2646 \\ \hline 3 & 1323 \\ \hline 3 & 441 \\ \hline 3 & 147 \\ \hline 7 & 49 \\ \hline 7 & 7 \\ \hline & 1 \\ \end{array}$

10584 = (2 × 2 × 2) × (3 × 3 × 3) × 7 × 7

Since 7 × 7 is left after grouping the prime factors in triplets, so 10584 cannot be expressed as the product of triplets of equal factors.

Hence, 10584 is not a perfect cube.

Find the smallest number by which 1323 must be multiplied so that the product is a perfect cube.

Answer

Resolving 1323 into prime factors, we get :

$\begin{array}{l|r} 3 & 1323 \\ \hline 3 & 441 \\ \hline 3 & 147 \\ \hline 7 & 49 \\ \hline 7 & 7 \\ \hline & 1 \\ \end{array}$

1323 = (3 × 3 × 3) × 7 × 7

Clearly, to make it a perfect cube it must be multiplied by 7.

⇒ 1323 × 7 = 9261

⇒ 9261 = (3 × 3 × 3) × (7 × 7 × 7), which is a perfect cube.

Hence, the smallest number by which 1323 must be multiplied so that the product is a perfect cube is 7.

What is the smallest number by which 1600 must be divided so that the quotient is a perfect cube?

Answer

Resolving 1600 into prime factors, we get :

$\begin{array}{l|r} 2 & 1600 \\ \hline 2 & 800 \\ \hline 2 & 400 \\ \hline 2 & 200 \\ \hline 2 & 100 \\ \hline 2 & 50 \\ \hline 5 & 25 \\ \hline 5 & 5 \\ \hline & 1 \\ \end{array}$

1600 = (2 × 2 × 2) × (2 × 2 × 2) × 5 × 5

Clearly, to make it a perfect cube, it must be divided by (5 × 5), i.e., 25.

⇒ 1600 ÷ 25 = 64

⇒ 64 = (2 × 2 × 2) × (2 × 2 × 2), which is a perfect cube.

Hence, the smallest number by which 1600 must be divided so that the quotient is a perfect cube is 25.

Find the smallest number by which 2560 must be multiplied so that the product is a perfect cube.

Answer

Resolving 2560 into prime factors, we get :

$\begin{array}{l|r} 2 & 2560 \\ \hline 2 & 1280 \\ \hline 2 & 640 \\ \hline 2 & 320 \\ \hline 2 & 160 \\ \hline 2 & 80 \\ \hline 2 & 40 \\ \hline 2 & 20 \\ \hline 2 & 10 \\ \hline 5 & 5 \\ \hline & 1 \\ \end{array}$

2560 = (2 × 2 × 2) × (2 × 2 × 2) × (2 × 2 × 2) × 5

Clearly, to make it a perfect cube it must be multiplied by (5 × 5), i.e., 25.

$\Rightarrow 2560 \times 25 = 64000\\[1em] \Rightarrow 64000 = (2 \times 2 \times 2) \times (2 \times 2 \times 2) \times (2 \times 2 \times 2) \times (5 \times 5 \times 5), \text{ which is a perfect cube.}$

Hence, the smallest number by which 2560 must be multiplied so that the product is a perfect cube is 25.

Find the cube root of:

$\dfrac{216}{2197}$

Answer

Solving,

$\Rightarrow \sqrt[3]{\dfrac{216}{2197}}\\[1em] \Rightarrow \dfrac{\sqrt[3]{216}}{\sqrt[3]{2197}}\\[1em] \Rightarrow \dfrac{\sqrt[3]{(2 \times 2 \times 2) \times (3 \times 3 \times 3)}}{\sqrt[3]{13 \times 13 \times 13}}\\[1em] \Rightarrow \dfrac{2 \times 3}{13}\\[1em] \Rightarrow \dfrac{6}{13}$

Hence, $\mathbf{\sqrt[3]{\dfrac{216}{2197}} = \dfrac{6}{13}}$.

Find the cube root of:

$4\dfrac{508}{1331}$

Answer

Solving,

$\Rightarrow \sqrt[3]{4\dfrac{508}{1331}}\\[1em] \Rightarrow \sqrt[3]{\dfrac{5832}{1331}}\\[1em] \Rightarrow \dfrac{\sqrt[3]{5832}}{\sqrt[3]{1331}}\\[1em] \Rightarrow \dfrac{\sqrt[3]{(2 \times 2 \times 2) \times (3 \times 3 \times 3) \times (3 \times 3 \times 3)}}{\sqrt[3]{11 \times 11 \times 11}}\\[1em] \Rightarrow \dfrac{2 \times 3 \times 3}{11}\\[1em] \Rightarrow \dfrac{18}{11}\\[1em] \Rightarrow 1\dfrac{7}{11}$

Hence, $\mathbf{\sqrt[3]{4\dfrac{508}{1331}} = 1\dfrac{7}{11}}$.

Find the cube root of:

42.875

Answer

Solving,

$\Rightarrow \sqrt[3]{42.875}\\[1em] \Rightarrow \sqrt[3]{\dfrac{42875}{1000}}\\[1em] \Rightarrow \dfrac{\sqrt[3]{42875}}{\sqrt[3]{1000}}\\[1em] \Rightarrow \dfrac{\sqrt[3]{(5 \times 5 \times 5) \times (7 \times 7 \times 7)}}{\sqrt[3]{10 \times 10 \times 10}}\\[1em] \Rightarrow \dfrac{5 \times 7}{10}\\[1em] \Rightarrow \dfrac{35}{10}\\[1em] \Rightarrow 3.5$

Hence, $\mathbf{\sqrt[3]{42.875} = 3.5}$.