Question

You are told that 1,331 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

1. 1331
Let us divide $1331$ into groups of three digits starting from the right. So $1331$ has two groups, one is $331$ and another is $1.$

For first group $331,$ digit $1$ is at one's place. $1$ comes at the unit's place of a number only when its cube root ends with $1.$ So unit's place of the required cube root is $1.$
For another group, i.e. $1,1^3=1$ and $2^3=8.$ So $1$ lies between $0$ and $8.$ The smaller number among $1$ and $2$ is $1.$ So the one's place of $1$ is $1$ and ten's place of cube root $1331$ is $1$
Hence $\sqrt[3]{31331}=11$

2. $4913$

Let us divide $4913$ into groups of three-digit starting from the right. So $4913$ has two groups one is $913$ and another is $4.$

For first group $913,$ the digit $3$ is at unit's place. $3$ comes at the unit's place of a number only when its cube root ends in $7.$ So unit's place of the required cube root is $7.$
For another group, i.e. $4,$ we know that $1^3=1$ and $2^3=8.$ We know that $4$ lies between $1$ and $8.$ The smaller number among $1$ and $2$ is $1.$ So the one's place of $1$ is $1$$andte{n}^{\prime }splaceofcuberoot$4913$is$1$Hence$\sqrt[3]{4913}=17$

3. $12167$

Let us divide $12167$ into groups of three-digit starting from the right. So $12167$ has two groups one is $167$ and another is $12.$

For first group $167,$ the digit $7$ is at unit's place. $7$ comes at a unit place of a number only when its cube root ends in $3.$ So unit's place of the required cube root is $3.$
For another group, i.e. $12,$ we know that $2^3=8$ and $3^3=27.$ We find that $12$ lies between $8$ and $27.$ The smaller number among $2$ and $3$ is $2.$ So the one's place of $2$ is $2$ itself and ten's place of cube root $12167$ is $2$
Hence $\sqrt[3]{312167}=23$

4. $32768$

Let us divide $32768$ into groups of three-digit starting from the right. So $32768$ has two groups one is $768$ and another is $32.$

For first group $768,$ the digit $8$ is at unit's place. $8$ comes at the unit place of a number only when its cube root ends in $2.$ So unit's place of the required cube root is $2.$
For another group, i.e. $32,3^3=27$ and $4^3=64.$ So $32$ lies between $27$ and $64.$ The smaller number among $3$ and $4$ is $4.$ So the ten's place of cube root $32768$ is $3$
Hence $\sqrt[3]{332768}=32$