Question

Find the cube root of each of the following numbers: 

(i) 8 x 125

(ii) -1728 x 216

(iii) -27 x 2744

(iv) -729 x -15625

Answer 1

(i) We know that for any two integers a and b, \(\sqrt[3]{ab}\) = \(\sqrt[3]{a}\times\sqrt[3]{b}\)

So from this property, we have:

\(\sqrt[3]{8\times125}\)

= \(\sqrt[3]{8}\times\sqrt[3]{125}\)

= \(\sqrt[3]{2\times2\times2}\times\sqrt[3]{5\times5\times5}\)

= \(2\times5\)

= 10

(ii) By Applying a and b \(\sqrt[3]{ab}\) =  \(\sqrt[3]{a}\times\sqrt[3]{b}\)

we have,

\(\sqrt[3]{-1728\times216}\)

= \(\sqrt[3]{-1728}\times\sqrt[3]{216}\)

= \(\sqrt[3]{-1728}\times\sqrt[3]{216}\)

To find out cube root by using units digit: 

Let’s take the number 1728. 

So, 

Unit digit = 8 

The unit digit in the cube root of 1728 = 2 

After striking out the units, tens and hundreds digits of the given number, we are left with the 1. As we know 1 is the largest number whose cube is less than or equals to 1. 

So, The tens digit of the cube root of 1728 = 1

∴ \(\sqrt[3]{1728}= 12\)

Prime factors of 216 = 2×2×2×3×3×3 

On grouping the factors in triples of equal factor, 

We have, 

216 = {2×2×2}×{3×3×3} 

Taking one factor from each group we get,

\(\sqrt[3]{216} = 2\times3 = 6\)

So,

\(\sqrt[3]{-1728\times216}\)

= \(\sqrt[3]{-1728}\times\sqrt[3]{216}\)

= \(-12\times6 = -72\)

(iii) By Applying a and b propertise, \(\sqrt[3]{ab}\) = \(\sqrt[3]{a}\times\sqrt[3]{b}\), we have,

\(\sqrt[3]{-27\times2744}\)

= \(\sqrt[3]{-27}\) x \(\sqrt[3]{2744}\)

= \(-\sqrt[3]{27}\) x \(\sqrt[3]{2744}\)

To find out cube root by using units digit: 

Let’s take the number 2744. 

So, 

Unit digit = 4 

The unit digit in the cube root of 2744= 4 

After striking out the units, tens and hundreds digits of the given number, we are left with the 2. 

As we know 1 is the largest number whose cube is less than or equals to 2. 

So, 

The tens digit of the cube root of 2744 = 1

∴ \(\sqrt[3]{2744}\) = 14

Prime factors of 216 = 2×2×2×3×3×3 

On grouping the factors in triples of equal factor, 

We have, 

216 = {2×2×2}×{3×3×3} 

Taking one factor from each group we get,

\(\sqrt[3]{216} \) = 2 x 3 = 6

So,

\(\sqrt[3]{-1728\times216}\)

= \(\sqrt[3]{-1728}\times\sqrt[3]{216}\)

= \(-12\times6 = -72\)

(iv) By Applying a and b properties, \(\sqrt[3]{ab}\) = \(\sqrt[3]{a}\times\sqrt[3]{b}\), We have

\(\sqrt[3]{-729\times-15625}\)

= \(\sqrt[3]{-729}\times\sqrt[3]{-15625}\)

= \(-\sqrt[3]{729}\) x \(-\sqrt[3]{15625}\)

To find out cube root by using units digit: 

Let’s take the number 15625. 

So, 

Unit digit = 5 

The unit digit in the cube root of 15625 = 5 

After striking out the units, tens and hundreds digits of the given number, we are left with the 15. 

As we know 2 is the largest number whose cube is less than or equals to 15(2³<15<3³). 

So, The tens digit of the cube root of 15625 = 2

∴ \(\sqrt[3]{15625} = 25\)

Also

\(\sqrt[3]{729}= 9,\)

As we know 9×9×9 = 729

Thus,

\(\sqrt[3]{-729\times-15625}\)

= \(\sqrt[3]{-729}\times\sqrt[3]{-15625}\)

= \(-9\times-256 = 225\)