Chapter 6.4, Problem 63HP

To determine

A number for which principal square-root and cube-root are both integers.

Answer to Problem 63HP

The number is 64.

Explanation of Solution

Given information: Consider the numbers from 1 to 10 an write their square and cubes.

Calculation: The square and cubes of first 10 non-zero numbers are written as,

  $\begin{array}{c}{\left(1\right)}^2=1{\left(1\right)}^3=1\\ {\left(2\right)}^2=4{\left(2\right)}^3=8\\ {\left(3\right)}^2=9{\left(3\right)}^3=27\\ {\left(4\right)}^2=16{\left(4\right)}^3=64\\ {\left(5\right)}^2=25{\left(5\right)}^3=125\\ {\left(6\right)}^2=36{\left(6\right)}^3=216\\ {\left(7\right)}^2=49{\left(7\right)}^3=343\\ {\left(8\right)}^2=64{\left(8\right)}^3=512\\ {\left(9\right)}^2=81{\left(9\right)}^3=729\\ {\left(10\right)}^2=100{\left(10\right)}^3=1000\end{array}$

Now, from squares and cubes the only number is 64 whose cube-root is 4 and square root is 8.

Hence, the only number is 64.