Cubes and Cube Roots — Check Your Progress

Show that the following number is a perfect cube. Also find the number whose cube is the given number:

74088

Answer

Finding prime factors of 74088 :

$\begin{array}{l|r} 2 & 74088 \\ \hline 2 & 37044 \\ \hline 2 & 18522 \\ \hline 3 & 9261 \\ \hline 3 & 3087 \\ \hline 3 & 1029 \\ \hline 7 & 343 \\ \hline 7 & 49 \\ \hline 7 & 7 \\ \hline & 1 \\ \end{array}$

74088 = (2 × 2 × 2) × (3 × 3 × 3) × (7 × 7 × 7)

Since 74088 can be expressed as the product of triplets of equal prime factors, so it is a perfect cube.

$\Rightarrow \sqrt[3]{74088}\\[1em] \Rightarrow \sqrt[3]{(2 \times 2 \times 2) \times (3 \times 3 \times 3) \times (7 \times 7 \times 7)}\\[1em] \Rightarrow 2 \times 3 \times 7\\[1em] \Rightarrow 42$

Hence, 74088 is a perfect cube and it is the cube of 42.

Show that the following number is a perfect cube. Also find the number whose cube is the given number:

15625

Answer

Finding prime factors of 15625 :

$\begin{array}{l|r} 5 & 15625 \\ \hline 5 & 3125 \\ \hline 5 & 625 \\ \hline 5 & 125 \\ \hline 5 & 25 \\ \hline 5 & 5 \\ \hline & 1 \\ \end{array}$

15625 = (5 × 5 × 5) × (5 × 5 × 5)

Since 15625 can be expressed as the product of triplets of equal prime factors, so it is a perfect cube.

$\Rightarrow \sqrt[3]{15625}\\[1em] \Rightarrow \sqrt[3]{(5 \times 5 \times 5) \times (5 \times 5 \times 5)}\\[1em] \Rightarrow 5 \times 5\\[1em] \Rightarrow 25$

Hence, 15625 is a perfect cube and it is the cube of 25.

Find the cube of the following number:

-17

Answer

Solving,

$\Rightarrow (-17)^3\\[1em] \Rightarrow (-17) \times (-17) \times (-17)\\[1em] \Rightarrow -4913$

Hence, $\mathbf{(-17)^3 = -4913}$.

Find the cube of the following number:

$-3\dfrac{4}{9}$

Answer

Solving,

$\Rightarrow \Big(-3\dfrac{4}{9}\Big)^3\\[1em] \Rightarrow \Big(\dfrac{-31}{9}\Big)^3\\[1em] \Rightarrow \dfrac{(-31) \times (-31) \times (-31)}{9 \times 9 \times 9}\\[1em] \Rightarrow \dfrac{-29791}{729}\\[1em] \Rightarrow -40\dfrac{631}{729}$

Hence, $\mathbf{\Big(-3\dfrac{4}{9}\Big)^3 = -40\dfrac{631}{729}}$.

Find the cube root of the following number by prime factorisation:

59319

Answer

Given number is 59319.

Finding prime factors of 59319 :

$\begin{array}{l|r} 3 & 59319 \\ \hline 3 & 19773 \\ \hline 3 & 6591 \\ \hline 13 & 2197 \\ \hline 13 & 169 \\ \hline 13 & 13 \\ \hline & 1 \\ \end{array}$

Expressing 59319 into prime factors, we have

59319 = (3 × 3 × 3) × (13 × 13 × 13)

$\Rightarrow \sqrt[3]{59319}\\[1em] \Rightarrow \sqrt[3]{(3 \times 3 \times 3) \times (13 \times 13 \times 13)}\\[1em] \Rightarrow 3 \times 13\\[1em] \Rightarrow 39$

Hence, cube root of 59319 = 39.

Find the cube root of the following number by prime factorisation:

21952

Answer

Given number is 21952.

Finding prime factors of 21952 :

$\begin{array}{l|r} 2 & 21952 \\ \hline 2 & 10976 \\ \hline 2 & 5488 \\ \hline 2 & 2744 \\ \hline 2 & 1372 \\ \hline 2 & 686 \\ \hline 7 & 343 \\ \hline 7 & 49 \\ \hline 7 & 7 \\ \hline & 1 \\ \end{array}$

Expressing 21952 into prime factors, we have

21952 = (2 × 2 × 2) × (2 × 2 × 2) × (7 × 7 × 7)

$\Rightarrow \sqrt[3]{21952}\\[1em] \Rightarrow \sqrt[3]{(2 \times 2 \times 2) \times (2 \times 2 \times 2) \times (7 \times 7 \times 7)}\\[1em] \Rightarrow 2 \times 2 \times 7\\[1em] \Rightarrow 28$

Hence, cube root of 21952 = 28.

Find the cube root of the following number:

-9261

Answer

Finding prime factors of 9261 :

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

Expressing 9261 into prime factors, we have

9261 = (3 × 3 × 3) × (7 × 7 × 7)

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

Hence, $\mathbf{\sqrt[3]{-9261} = -21}$.

Find the cube root of the following number:

$2\dfrac{43}{343}$

Answer

Converting mixed fraction to improper fraction:

$2\dfrac{43}{343} = \dfrac{729}{343}$

$\Rightarrow \sqrt[3]{2\dfrac{43}{343}}\\[1em] \Rightarrow \sqrt[3]{\dfrac{729}{343}}\\[1em] \Rightarrow \dfrac{\sqrt[3]{729}}{\sqrt[3]{343}}\\[1em] \Rightarrow \dfrac{\sqrt[3]{(3 \times 3 \times 3) \times (3 \times 3 \times 3)}}{\sqrt[3]{7 \times 7 \times 7}}\\[1em] \Rightarrow \dfrac{3 \times 3}{7}\\[1em] \Rightarrow \dfrac{9}{7}\\[1em] \Rightarrow 1\dfrac{2}{7}$

Hence, $\mathbf{\sqrt[3]{2\dfrac{43}{343}} = 1\dfrac{2}{7}}$.

Find the cube root of the following number:

0.216

Answer

Converting decimal to fraction:

$0.216 = \dfrac{216}{1000}$

$\Rightarrow \sqrt[3]{0.216}\\[1em] \Rightarrow \sqrt[3]{\dfrac{216}{1000}}\\[1em] \Rightarrow \dfrac{\sqrt[3]{216}}{\sqrt[3]{1000}}\\[1em] \Rightarrow \dfrac{\sqrt[3]{(2 \times 2 \times 2) \times (3 \times 3 \times 3)}}{\sqrt[3]{10 \times 10 \times 10}}\\[1em] \Rightarrow \dfrac{2 \times 3}{10}\\[1em] \Rightarrow \dfrac{6}{10}\\[1em] \Rightarrow 0.6$

Hence, $\mathbf{\sqrt[3]{0.216} = 0.6}$.

Find the smallest number by which 5184 should be multiplied so that product is a perfect cube. Also find the cube root of the product.

Answer

Finding prime factors of 5184 :

$\begin{array}{l|r} 2 & 5184 \\ \hline 2 & 2592 \\ \hline 2 & 1296 \\ \hline 2 & 648 \\ \hline 2 & 324 \\ \hline 2 & 162 \\ \hline 3 & 81 \\ \hline 3 & 27 \\ \hline 3 & 9 \\ \hline 3 & 3 \\ \hline & 1 \\ \end{array}$

5184 = (2 × 2 × 2) × (2 × 2 × 2) × (3 × 3 × 3) × 3

Since 3 is left after grouping the prime factors in triplets, so two more 3's are needed to complete the triplet.

∴ The required number = 3 × 3 = 9

Product = 5184 × 9 = 46656

$\Rightarrow \sqrt[3]{46656}\\[1em] \Rightarrow \sqrt[3]{(2 \times 2 \times 2) \times (2 \times 2 \times 2) \times (3 \times 3 \times 3) \times (3 \times 3 \times 3)}\\[1em] \Rightarrow 2 \times 2 \times 3 \times 3\\[1em] \Rightarrow 36$

Hence, 5184 should be multiplied by 9 to get the perfect cube 46656 and $\mathbf{\sqrt[3]{46656} = 36}$.

Find the smallest number by which 8788 should be divided so that quotient is a perfect cube. Also find the cube root of the quotient.

Answer

Finding prime factors of 8788 :

$\begin{array}{l|r} 2 & 8788 \\ \hline 2 & 4394 \\ \hline 13 & 2197 \\ \hline 13 & 169 \\ \hline 13 & 13 \\ \hline & 1 \\ \end{array}$

8788 = 2 × 2 × (13 × 13 × 13)

Since 2 × 2 is left after grouping the prime factors in triplets, so 8788 should be divided by 2 × 2 = 4.

Quotient = 8788 ÷ 4 = 2197

$\Rightarrow \sqrt[3]{2197}\\[1em] \Rightarrow \sqrt[3]{13 \times 13 \times 13}\\[1em] \Rightarrow 13$

Hence, 8788 should be divided by 4 to get the perfect cube 2197 and $\mathbf{\sqrt[3]{2197} = 13}$.

Find the side of a cube whose volume is 4096 cm³.

Answer

Let the length of the side of the cube be $a$ cm.

We know that, volume of a cube = (side)³

Finding prime factors of 4096 :

$\begin{array}{l|r} 2 & 4096 \\ \hline 2 & 2048 \\ \hline 2 & 1024 \\ \hline 2 & 512 \\ \hline 2 & 256 \\ \hline 2 & 128 \\ \hline 2 & 64 \\ \hline 2 & 32 \\ \hline 2 & 16 \\ \hline 2 & 8 \\ \hline 2 & 4 \\ \hline 2 & 2 \\ \hline & 1 \\ \end{array}$

Expressing 4096 into prime factors, we have

4096 = (2 × 2 × 2) × (2 × 2 × 2) × (2 × 2 × 2) × (2 × 2 × 2)

$\Rightarrow a^3 = 4096\\[1em] \Rightarrow a = \sqrt[3]{4096}\\[1em] \Rightarrow a = \sqrt[3]{(2 \times 2 \times 2) \times (2 \times 2 \times 2) \times (2 \times 2 \times 2) \times (2 \times 2 \times 2)}\\[1em] \Rightarrow a = 2 \times 2 \times 2 \times 2\\[1em] \Rightarrow a = 16$

Hence, the side of the cube is 16 cm.

Find the volume of a cubical box if the cost of painting its outer surface is ₹1080 at the rate of ₹45 per sq. m.

Answer

Total cost of painting = ₹1080

Rate of painting = ₹45 per sq. m

$\Rightarrow \text{Outer surface area} = \dfrac{\text{Total cost}}{\text{Rate}}\\[1em] = \dfrac{1080}{45}\\[1em] = 24 \text{ sq. m}$

Let the length of the side of the cubical box be $a$ m.

Surface area of the cubical box = $6a^2$

$\Rightarrow 6a^2 = 24\\[1em] \Rightarrow a^2 = \dfrac{24}{6}\\[1em] \Rightarrow a^2 = 4\\[1em] \Rightarrow a = \sqrt{4}\\[1em] \Rightarrow a = 2$

Volume of the cubical box = $a^3$ = 2 × 2 × 2 = 8

Hence, the volume of the cubical box is 8 m³.

In a Maths lab there are some cubes and cuboids of the following measurements:

(i) one cube of side 4 cm

(ii) one cube of side 6 cm

(iii) 3 cuboids each of dimensions 4 cm × 4 cm × 6 cm and 3 cuboids each of dimensions 4 cm × 6 cm × 6 cm.

A student wants to arrange these cubes and cuboids in the form of a big cube. Is it possible for him/her to arrange them in the form of a big cube? If yes, then find the length of side of new cube so formed.

Answer

Volume of the cube of side 4 cm = 4 × 4 × 4 = 64 cm³

Volume of the cube of side 6 cm = 6 × 6 × 6 = 216 cm³

Volume of 3 cuboids of dimensions 4 cm × 4 cm × 6 cm = 3 × (4 × 4 × 6) = 3 × 96 = 288 cm³

Volume of 3 cuboids of dimensions 4 cm × 6 cm × 6 cm = 3 × (4 × 6 × 6) = 3 × 144 = 432 cm³

Total volume = 64 + 216 + 288 + 432 = 1000 cm³

Let the length of the side of the new cube be $a$ cm.

$\Rightarrow a^3 = 1000\\[1em] \Rightarrow a = \sqrt[3]{1000}\\[1em] \Rightarrow a = \sqrt[3]{10 \times 10 \times 10}\\[1em] \Rightarrow a = 10$

Since 1000 is a perfect cube, so the given cubes and cuboids can be arranged in the form of a big cube.

Hence, yes, a big cube can be formed and the length of its side is 10 cm.