[Math] Find the coefficient of $x^{-2}$ in the expansion of $(x-1)^3(\frac{1}{x} +2x)^6$

algebra-precalculusbinomial theorembinomial-coefficients

Find the coefficient of $x^{-2}$ in the expansion of $(x-1)^3(\frac{1}{x} +2x)^6$.

I have tried finding out the general term of the whole expression.

Then, I get the result as such
$$[(-1)^r {3 \choose r}(x^{3-r})][{6 \choose r} (2x)^{6-r}(\frac{1}{x})^r]$$
But after solving for I get $r = 11/3$

I think I am doing it wrong by just multiplying the general term. I don't know how else to continue. Please help.

Thank You!

Edit : There was a typo in the question

Best Answer

We have \begin{align}(x-1)^3(\frac1x+2x)^6&=\left(\frac1x\right)^6(x-1)^3(1+2x^2)^6\\ &=\frac1{x^6}\left(\sum_{r=0}^3\binom3rx^{3-r}(-1)^r\right)\left(\sum_{s=0}^6\binom6s(2x^2)^s\right)\\ \end{align}

In order to get an exponent of $-2$, we can ignore the $\dfrac1{x^6}$ and look for an exponent of $4$ instead. Therefore, we have $r=1$ and $s=1$, OR $r=3$ and $s=2$.

As such, we have $$\binom31(-1)^1\binom61(2)^1+\binom33(-1)^3\binom62(2)^2=\boxed{-96}.$$

Checking with WolframAlpha confirms the answer.

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