help with wrong result for $v(x) = \frac{\sqrt[3]{x-1}}{(x+2)^2}$ I need to differentiate this:
$$v(x) = \frac{\sqrt[3]{x-1}}{(x+2)^2}$$
I used this formula:
$$ \frac{f'(x)g(x) - f(x)g'(x)}{g^2(x)}$$
Where:
$$ f'(x) = \frac{1}{3\sqrt[3]{(x-1)^2}}$$
and
$$ g'(x) = 2(x+2)$$
so my result is:
$$ \frac{ \frac{1}{3\sqrt[3]{(x-1)^2}}(x+2)^2 - \sqrt[3]{(x-1)} \cdot2(x+2)}{(x+2)^4} = \frac{ \frac{1}{3\sqrt[3]{(x-1)^2}}(x+2)^2 -  2\sqrt[3]{(x-1)}(x+2) }{(x+2)^4} = \frac{ (x+2)^2 -  2\sqrt[3]{(x-1)}(x+2) }{3\sqrt[3]{(x-1)^2}(x+2)^4} = \frac{ (x+2) \big[ (x+2) -  2\sqrt[3]{(x-1)} \big] }{3\sqrt[3]{(x-1)^2}(x+2)^4} = \frac{ (x+2) -  2\sqrt[3]{(x-1)} }{3\sqrt[3]{(x-1)^2}(x+2)^3}$$
but according to wolframalpha the result should be:
$$\frac{ 8 - 5x }{3\sqrt[3]{(x-1)^2}(x+2)^3}$$

Multiplying top and bottom of 
$$\frac{\frac{1}{3\sqrt[3]{(x-1)^2}}(x+2)^2-2\sqrt[3]{x-1}\ (x+2)}{(x+2)^4}$$
by $3\sqrt[3]{(x-1)^2}$ gives you
$$\frac{(x+2)^2-2\sqrt[3]{x-1}\ (x+2)\cdot\color{red}{3\sqrt[3]{(x-1)^2}}}{3\sqrt[3]{(x-1)^2}(x+2)^4}.$$
