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

Question

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 $-3$.

Answer 1

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, $a = -3$.

Answer 2

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

Answer 3

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

Answer 4

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 $?$