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:

$3^{-8}{ 10 \choose 8 }$

  • $\begingroup$ Yeah, where did the negative sign go? $\endgroup$ Jan 15, 2017 at 18:12
  • $\begingroup$ Your answer looks right to me. What's the source of the solution? (For $k=8$ the minus sign will not matter, but where did the $3$ come from?) $\endgroup$ Jan 15, 2017 at 18:16
  • 3
    $\begingroup$ Is there perhaps a typo: we have $\left(\frac{1}{x^3}\right)^{10-k} = \left(\frac{1}{x}\right)^{3(10-k)}$ so if it's supposed to be $1/x$ instead of $1/x^3$ inside the bracket then it becomes correct. $\endgroup$
    – Winther
    Jan 15, 2017 at 18:18
  • $\begingroup$ @Winther I think so. And one typo in the "solution" makes a second (the missing minus sign) more likely. $\endgroup$ Jan 15, 2017 at 18:23
  • $\begingroup$ Yes @Winther is definitely right, thanks. But it wasn't easy for me to see cause I've spent ages on this problem. $\endgroup$ Jan 15, 2017 at 18:40

1 Answer 1


$$\begin{align} \left(\frac x3-\frac 1{x^3}\right)^{10} &=\frac {x^{10}}{3^{10}}\left(1-\frac 3{x^4}\right)^{10}\\ &=\frac {x^{10}}{3^{10}}\sum_{r=0}^{10}\binom {10}r\left(- 3x^{-4}\right)^r \end{align}$$ Putting $r=2$ gives $x^2$ term, hence coefficient of $x^2$ is $$\frac 1{3^{10}} \binom{10}2(-3)^2=\frac {45}{3^8}=\color{red}{\frac 5{3^6}}$$


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