Question

If $72K$ is a perfect cube, find the value of $K$.

• A. $1$
• B. $2$
• C. $3$
• D. $4$

Answer 1

Answer: (C)

Given : $72K=x^3$

Let us prime factorize 72 .

$72=2\times 2\times 2\times 3\times 3$..(i)

For a number to be a perfect cube , it must have all the numbers in triplet.

Here , 3 is not in the form of triplet, hence we multiply (i) both the sides with 3.

$\Rightarrow 72\times 3=2\times 2\times 2\times 3\times 3\times 3$

$\Rightarrow 216=2\times 2\times 2\times 3\times 3\times 3$

Now 2 and 3 are in triplets.

$\Rightarrow 72K=216$

$\Rightarrow K=\frac{216}{72}$

$\Rightarrow K=3$

The value of $K$ is $3$