Problem with binomial expansion

Question

So here is the problem:

Find the constant term in the expansion of (x-(2/x))^2 ·（x^2 +(2/x))^3

I understand I can just use my calculator to figure out the answer, but is there any simple way to solve that ? Thanks!

Answer 1

Here is a proposal which might be convenient.

• At first note that it is helpful to have the first rows of Pascal's triangle in mind at least $1;\quad 1,1;\quad 1,2,1;\quad 1,3,3,1$.

• So, we already know $(a+b)^2=a^2+2ab+b^2$ and $(a+b)^3=a^3+3a^2b+3ab^2+b^3$.

Here we also use the coefficient of operator $[x^n]$ to denote the coefficient of $x^n$ just for ease of notation.

Having these prerequisits close at hand, we can calculate \begin{align*} [x^0]&\left(x-\frac{2}{x}\right)^2\left(x^2+\frac{2}{x}\right)^3\\ &=[x^0]\left(x^2-4+\frac{4}{x^2}\right)\left(x^6+6x^3+12+\frac{8}{x^3}\right)\tag{1}\\ &\,\,\color{blue}{=-48}\tag{2} \end{align*}

Comment:

• In (1) we apply the binomial theorem twice and expand the binomials.

• In (2) we take a closer look at the left-hand factor $\left(x^2-4+\frac{4}{x^2}\right)$.

  • Looking at $x^2$ we need a term with $\frac{1}{x^2}$ at the right-hand factor, which don't exist.

  • Looking at $-4$ we see we can take $12$ at the right-hand factor, which gives $-48$.

  • Looking at $\frac{4}{x^2}$ we need a term with $x^2$ at the right-hand factor, which don't exist.

  • No further considerations are necessary.