Find the exponent for $x^2$ in the expansion of:
$(\frac{x}{3} - \frac{1}{x^3})^{10}$
What trips me up is that the second term is negative. Though I find using the binomial theorem that the expansion should be a sum of terms on the form:
${ 10 \choose k}(\frac{x}{3})^k(-\frac{1}{x^3})^{10-k} $
But according to the solution, that is wrong. The form is supposed to be:
${ 10 \choose k}(\frac{x}{3})^k(\frac{1}{x^3})^{3(10-k)} $
A solution is then found for $k = 8$ which is simple. But I don't understand why my form is wrong and the solutions form is right. Why is the coefficient for the second factor $3(10-k)$ and where did the negative sign go?
The solution for the problem is supposed to be:
$3^{-8}{ 10 \choose 8 }$
$$\begin{align} \left(\frac x3-\frac 1{x^3}\right)^{10} &=\frac {x^{10}}{3^{10}}\left(1-\frac 3{x^4}\right)^{10}\\ &=\frac {x^{10}}{3^{10}}\sum_{r=0}^{10}\binom {10}r\left(- 3x^{-4}\right)^r \end{align}$$ Putting $r=2$ gives $x^2$ term, hence coefficient of $x^2$ is $$\frac 1{3^{10}} \binom{10}2(-3)^2=\frac {45}{3^8}=\color{red}{\frac 5{3^6}}$$