Question

If a positive rational number n is not a perfect cube of a rational number then n² is also not a perfect cube of a rational number.

Answer 1

$$\begin{gathered} Giventhatnisnotperfectcubeofarationalnumber\frac{a}{b}\left(say\right) \\ Then \\ nnotequalto{\left(\frac{a}{b}\right)}^3 \\ Cubing \\ {n}^{3}notequalto{\left({\left(\frac{a}{b}\right)}^3\right)}^3 \\ {n}^{3}notequalto{\left(\frac{a^3}{b^3}\right)}^3 \\ {n}^{3}notequalto{\left(\frac{p}{q}\right)}^{3}wherep=a^{3}andq=b^3 \\ So{n}^{3}isalsonotperfectcubeofarationalnumber \end{gathered}$$