Question

If $x^{-2}=64,then{x}^{\frac{1}{3}}+x^0=$

• A. 2
• B. 3
• C. $\frac{3}{2}$
• D. $\frac{2}{3}$

Answer 1

The correct option is C

$\frac{3}{2}$

$x^{-2}=64\Rightarrow \frac{1}{x^2}=64\Rightarrow$ ${\left(\frac{1}{x}\right)}^2=(8{)}^2\therefore \frac{1}{x}=8\Rightarrow x=\frac{1}{8}{x}^{\frac{1}{3}}+x^0={\left(\frac{1}{8}\right)}^{\frac{1}{3}}+1={\left[{\left(\frac{1}{2}\right)}^3\right]}^{\frac{1}{3}}+1={\left(\frac{1}{2}\right)}^{3\times \frac{1}{3}}+1=\frac{1}{2}+1=\frac{3}{2}$