Question

Find out if $10648$ is a perfect cube or not.

Answer 1

Hint: Split $10648$ into factors of $2$ and $11$. Then, first divide $10648$ by $2$ until we get the number which is not divisible by $2$. Then, continue with 11 till it cannot be divided any further. Now, collect the same factors in a group of 3 to get the cube of the number, then express the factors in the exponents of $3$.

Complete step by step answer:
Before proceeding with the question, we must know how to find the cube of the number. The cube of a number can be found out by finding the factors and then arranging the same factors in a group of 3.

In this question we have to find out if $10648$ is a perfect cube or not. $10648$ is a perfect cube or not can be found out by finding the factors of $10648$.

First, we can divide $10648$ by $2$ until we get a number which cannot be divided by $2$.

After dividing $10648$ by $2$ we get,

$10648\div 2=5324$

Again dividing $5324$ by $2$ we get,

$5324\div 2=2662$

Again dividing $2662$ by $2$we get,

$2662\div 2=1331$

Now, $1331$ cannot be divided by 2. We have to find the number through which it gets divided and that number is $11$.

Therefore, dividing $1331$ by $11$ we get,

$1331\div 11=121$

Again dividing $121$ by $11$ we get,

$121\div 11=11$

Again dividing $11$ by $11$ we get,

$11\div 11=1$

Now, we can collect the factors for $10648$ and we can write it as,

$10648=2\times 2\times 2\times 11\times 11\times 11$

Now, expressing each factor in exponents of three we get,

$\Rightarrow 10648={{2}^{3}}\times {{11}^{3}}$

$\Rightarrow \sqrt[3]{10468}=\sqrt[3]{{{2}^{3}}}\times \sqrt[3]{{{11}^{3}}}$

$\Rightarrow \sqrt[3]{10648}=2\times 11$

$\therefore \sqrt[3]{10648}=22$

Hence, $10648$ is a perfect cube.


Note: Be careful to group the factors in the exponents of $3$ for finding the cube root of any number. The mistake that can be made here is by reading the question wrongly and taking the factors in the exponents of 2 instead of 3 and writing the wrong answer.