rational exponent of negative base

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 $(-8)^{2/3}$

• notice that $(-8)^{2/3}$ is a complex number Commented Oct 26, 2014 at 22:20

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 $4, -2 - 2i\sqrt{3}, -2 + 2i\sqrt{3}$.

• Thanks, sorry if I'm missing something obvious. I see how (-8)^(2/3) has multiple roots but so does (8)^(2/3) so why do I get an error on my calculator when the 8 is negative. Commented Oct 27, 2014 at 4:36
• When I enter (-8)^(2/3) on my calculator and the calculator on my computer, they both give me 4. I would get an error on my calculator if I used a subtraction sign instead of a negative sign, otherwise I cannot think of a reason (other than improper programming) why your calculator is giving you an error when you try to compute $(-8)^{\frac{2}{3}}$. Commented Oct 27, 2014 at 10:20
• I never actually noticed there was a separate negative button but on my fx-300MS I get a "Math ERROR" either way I input it. Commented Oct 27, 2014 at 19:16
• Because the function $a^x$ with real $x$ is only defined for positive $a$. You might want to try the cube root button instead (if there is one), or write $8^{2/3}$ instead since they mean the same (in the reals), as $(-8)^{1/3} = -8^{1/3}$ Commented Oct 30, 2014 at 0:13