Question

You are told that $1331$ is a perfect cube. Can you guess without factorisation what is its cube root? Similarly guess the cube roots of $4913,12167,32768$

Answer 1

Step 1: Find the cube root of $1331$.

Make group of three digits

Unit digit of cube root of $1331=$Unit digit of cube root of first group $331$

Since, $331$ ends in $1$, so unit digit of cube root of $331$ is also $1$.

Thus, unit digit of cube root of $1331$ is $1$.

Now for the second group we can see that $1^3=1$

Therefore, cube root of $1331$ is $11$.

Step 2: Find the cube root of $4913$.

Make group of three digits

Unit digit of cube root of $4913=$Unit digit of cube root of first group $913$

Since, $913$ ends in $3$, so unit digit of cube root of $913$ is $7$.

Thus, unit digit of cube root of $4913$ is $7$.

Now for the second group $4$

We know that $1^3=1$and $2^3=8$

And $1<4<8$

So, we will choose the smaller number as the tens digit of the cube root, i.e., $1$

Therefore, cube root of $4913$ is $17$.

Step 3: Find the cube root of $12167$

Make group of three digits

Unit digit of cube root of $12167=$Unit digit of cube root of first group $167$

Since, $167$ ends in $7$, so unit digit of cube root of $167$ is $3$.

Thus, unit digit of cube root of $12167$ is $3$.

Now for the second group $12$

We know that $2^3=8$and $3^3=27$

And $8<12<27$

So, we will choose the smaller number as the tens digit of cube root, i.e., $2$

Therefore, cube root of $12167$ is $23$.

Step 4: Find the cube root of $32768$

Make group of three digits

Unit digit of cube root of $32768=$Unit digit of cube root of first group $768$

Since, $768$ ends in $8$, so unit digit of cube root of $768$ is $2$.

Thus, unit digit of cube root of $32768$ is $2$.

Now for the second group $32$

We know that $3^3=27$and $4^3=64$

And $27<32<64$

So, we will choose the smaller number as the tens digit of the cube root, i.e., $3$

Therefore, cube root of $32768$ is $32$.

Hence, the cube root of $1331,4913,12167$ and $32768$ are $11,17,23$ and $32$ respectively.