Question

If $x^{2/3}-7x^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}+10=0$, then the value of $x$ is:

• A. $125$
• B. $8$
• C. $\varnothing$
• D. $125,8$

Answer 1

The correct option is D

$125,8$

Explanation for the correct option.

In $x^{2/3}-7x^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}+10=0$,

Let $x^{1/3}=y$, then $x^{2/3}-7x^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}+10=0$, can be written as

$\begin{array}{rcl}{y}^2-7y+10& =& 0\\ & \Rightarrow & \left(y-2\right)\left(y-5\right)=0\\ \Rightarrow y& =& 2\mathrm{or}5\end{array}$

This means $x^{1/3}=2\mathrm{or}5$.

Therefore, $x=2^{3}or{5}^3=\left\{8,125\right\}$

Hence, option D is correct.