Help me solve this And please explain
\((-2)^3=-8\\ \left(\frac{3}{4}\right)^2=\frac{9}{16}\)
\((-2)^3=(-2)\times(-2)\times(-2)=-8\\\\\\\left(\frac{3}{4}\right)^2=\frac{3^2}{4^2}=\frac{3\times3}{4\times4}=\frac{9}{16}\)
