I have the definite integral $$\int_{1}^{\,9} {\frac{6}{\sqrt[3]{x-9}}}\, \mathrm dx$$
When I try to evaluate it I get the indefinite integral equals $9(x-9)^{2/3}$ and evaluating at the limits gives me $0-9(-8)^{2/3}$
When I calculate $(-8)^{2/3}$ my calculator gives me an error but Mathematica says it's $4(-1)^{2/3}$ or the complex number $-2+2i\sqrt{3}$
If I try to do it as $((-8)^{1/3})^2$ my calculator gives me 4 but Mathematica gives the same complex number it calculated for $(-8)^{2/3}$
$((-8)^2)^{1/3}$ both my calculator and Mathematica give the answer = 4
My homework question was to determine if the integral was divergent otherwise to evaluate the integral. The correct solution was when I chose $(-8)^{2/3} = 4$ and therefore $$\int_{1}^{\,9} {\frac{6}{\sqrt[3]{x-9}}}\,\mathrm dx = -36$$
Can someone explain why I got different answers for $(-8)^{2/3}$
If you want to find the real root, use the definition of rational roots to obtain
$$(-8)^{\frac{2}{3}} = [(-8)^{\frac{1}{3}}]^2 = (\sqrt[3]{8})^2 = (-2)^2 = 4$$
or, equivalently,
$$(-8)^{\frac{2}{3}} = [(-8)^2]^{\frac{1}{3}} = 64^{\frac{1}{3}} = \sqrt[3]{64} = 4$$
However, $-8$ has three complex cube roots, one of which is $-2$. You can find them by using DeMoivre's Theorem or solving the equation $x^3 + 8 = 0$.
\begin{align*} x^3 + 8 & = 0\\ (x + 2)(x^2 + 2x + 4) & = 0 \end{align*}
Setting each factor equal to zero yields \begin{align*} x + 2 & = 0 & x^2 + 2x + 4 & = 0\\ x & = -2 & x^2 + 2x & = -4\\ & & x^2 + 2x + 1 & = -3\\ & & (x + 1)^2 & = -3\\ & & x + 1 & = \pm\sqrt{-3}\\ & & x + 1 & = \pm i\sqrt{3}\\ & & x & = -1 \pm i\sqrt{3} \end{align*}
If we square $(-8)^{1/3}$, we obtain $(-8)^{\frac{2}{3}}$. Squaring $-2$ yields $4$. If we square $-1 + i\sqrt{3}$, we obtain
\begin{align*} (-1 + i\sqrt{3})^2 & = 1 - 2i\sqrt{3} + 3i^2\\ & = 1 - 2i\sqrt{3} - 3\\ & = -2 - 2i\sqrt{3} \end{align*}
Squaring $-1 - i\sqrt{3}$ yields $-2 + 2i\sqrt{3}$. Hence, the complex values of $(-8)^{\frac{2}{3}}$ are $4, -2 - 2i\sqrt{3}, -2 + 2i\sqrt{3}$.