<s>[INST] Trouble with factoring polynomial fractions and understanding wolfram alpha I hope the community can excuse me if I'm making excessive posts, I have a calculus quiz tomorrow and I want to be as prepared as possible.
I'm going through a review problem on wolfram alpha and I have trouble understanding a step. The problem begins like this:
$$\frac{2}{3}x^{\frac{-1}{3}}(x-5) + x^{\frac{2}{3}} = 0$$
Wolfram alpha combines and rewrites as follows:
$$\frac{2(x - 5)}{3\sqrt[3]{x}} + x^{\frac{2}{3}} = 0$$
So far so good. But now wolfram alpha subtracts $x^\frac{2}{3}$ from both sides...
$$\frac{2x^{\frac{2}{3}}}{3} – \frac{10}{3(\sqrt[3]{x})} = -x^{\frac{2}{3}}$$
And that's my problem, right there—why is there a three under the $\frac{2x^{\frac{2}{3}}}{3}$? If wolfram alpha multiplied out 2(x - 5) then wouldn't the resulting numbers share a denominator of $3(\sqrt[3]{x})$?
 [/INST] Let's write radicals in fractional notation:
$$\frac{2(x - 5)}{3\sqrt[3]{x}} + x^{2/3} = 0\iff \frac{2(x - 5)}{3x^{1/3}} + x^{2/3}=0.$$
Now, it will become obvious to you: $$\require{cancel}\eqalign{\frac{2(x - 5)}{3x^{1/3}} + x^{2/3}
&=\frac{2x - 10}{3x^{1/3}} + x^{2/3}\\ \ \\
&=\frac{2x}{3x^{1/3}} - \frac{10}{3x^{1/3}} + x^{2/3}\\ \ \\
&=\frac{2x^{1/3+2/3}}{3x^{1/3}} - \frac{10}{3x^{1/3}} + x^{2/3}\\ \ \\
&=\frac{2x^{1/3}\cdot x^{2/3}}{3x^{1/3}} - \frac{10}{3x^{1/3}} + x^{2/3}\\ \ \\
&=\frac{2\cancel{x^{1/3}}\cdot x^{2/3}}{3\cancel{x^{1/3}}} - \frac{10}{3x^{1/3}} + x^{2/3}\\ \ \\
&=\frac{{2}x^{2/3}}{3} - \frac{10}{3x^{1/3}} + x^{2/3}.
}$$
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