# Problem with binomial expansion

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!

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.