# $(x^3 + \frac{a}{x^2})^5 = -270$, find $a$

I am working on my scholarship exam practice and not sure how to begin. Please assume math knowledge at high school or pre-university level.

Let $$a$$ be a real constant. If the constant term of $$(x^3 + \frac{a}{x^2})^5$$ is equal to $$-270$$, then $$a=$$......

Could you please give a hint for this question? The answer provided is $$-3$$.

From the binomial theorem, we know that the sum of the powers of $$x^3$$ and $$a/x^2$$ must add to $$5$$. So if our term in the expansion is $$k (x^3)^p (\frac{a}{x^2})^q$$, $$p+q = 5$$. Moreover, we want our term to be constant (no nonzero powers of $$x$$), so we have $$3p - 2q = 0$$. Solving these two expressions, we have $$p = 2, q =3$$. So we must find $$a$$ such that the coefficient of the $$(x^3)^2 (\frac{a}{x^2})^3$$ term is $$270$$.

The binomial theorem says that the coefficient of this power in the expansion is $${5 \choose 2} = 10$$, so in particular, the coefficient of the constant term is $$10a^3$$, which we want to equal $$270$$. Therefore, $$a = -3$$.

Hint: binomial expand $$(x^3+ax^{-2})^5$$.

Hint: $$(A+B)^5={A}^{5}+5\,{A}^{4}B+10\,{A}^{3}{B}^{2}+10\,{A}^{2}{B}^{3}+5\,A{B}^{4}+ {B}^{5}$$

Hint:

The general term $$T_{r+1}$$ is $$\binom5r(x^3)^{5-r}\left(\dfrac a{x^2}\right)^r=\binom5ra^rx^{3\cdot5-3r-2r}$$

For the constant term, the exponent of $$x$$ will be $$?$$