Question

Solve the given equation.

$x^{\frac{2}{3}}-{10}^{\frac{1}{3}}+24=0$

Short answer

The solutions are $x=64$and$x=216$.

Step-by-step solution

Step 1. Given information.

The given equation is$x^{\frac{2}{3}}-{10}^{\frac{1}{3}}+24=0$.

Step 2. Simplify the equation.

$\begin{array}{rcl}{x}^{\frac{2}{3}}-10x^{\frac{1}{3}}+24& =& 0\\ {\left(x^{\frac{1}{3}}\right)}^2-10x^{\frac{1}{3}}+24& =& 0\\ {\left(x^{\frac{1}{3}}\right)}^2-6x^{\frac{1}{3}}-4x^{\frac{1}{3}}+24& =& 0\\ x^{\frac{1}{3}}\left(x^{\frac{1}{3}}-6\right)-4\left(x^{\frac{1}{3}}-6\right)& =& 0\\ \left(x^{\frac{1}{3}}-4\right)\left(x^{\frac{1}{3}}-6\right)& =& 0\end{array}$

Step 3. Solve the equation for x.

$\begin{array}{rcl}{x}^{\frac{1}{3}}-6& =& 0\\ x^{\frac{1}{3}}& =& 6\\ {\left(x^{\frac{1}{3}}\right)}^3& =& 6^3\\ x& =& 216\end{array}$ and $\begin{array}{rcl}{x}^{\frac{1}{3}}-4& =& 0\\ x^{\frac{1}{3}}& =& 4\\ {\left(x^{\frac{1}{3}}\right)}^3& =& 4^3\\ x& =& 64\end{array}$

Step 4. Simplified answer.

Hence, the required solutions are$x=64$and$x=216$.