Question

Find the cube root of the following number −1728 × 216 .

Answer 1

Property:
For any two integers a and b,

\[\sqrt[3]{ab} = \sqrt[3]{a} \times \sqrt[3]{b}\]

From the above property, we have:​

\[\sqrt[3]{- 1728 \times 216}\]

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

\[= - \sqrt[3]{1728} \times \sqrt[3]{216}\]     (For any positive integer x, \[\sqrt[3]{- x} = - \sqrt[3]{x}\]

Cube root using units digit:
Let us consider the number 1728.
The unit digit is 8; therefore, the unit digit in the cube root of 1728 will be 2.
After striking out the units, tens and hundreds digits of the given number, we are left with 1.
Now, 1 is the largest number whose cube is less than or equal to 1.
Therefore, the tens digit of the cube root of 1728 is 1.

∴ \[\sqrt[3]{1728} = 12\]

On factorising 216 into prime factors, we get:

\[216 = 2 \times 2 \times 2 \times 3 \times 3 \times 3\]

On grouping the factors in triples of equal factors, we get:

\[216 = \left\{ 2 \times 2 \times 2 \right\} \times \left\{ 3 \times 3 \times 3 \right\}\]

Now, taking one factor from each triple, we get:

\[\sqrt[3]{216} = 2 \times 3 = 6\]

Thus

\[\sqrt[3]{- 1728 \times 216} = - \sqrt[3]{1728} \times \sqrt[3]{216} = - 12 \times 6 = - 72\]