How to solve an equation with a power of 5?

Question

Let $a$ be a real constant, if $$\left(x^3 + \frac{a}{x^2}\right)^5 = -270$$what is the value of $a$ ?

Is this a type of binomial expansion question?

Answer 1

Hint.

By using binomial expansion, the formula is $(a+b)^5 = a^5 + 5a^4b + 10a^3b^2 + 10a^2b^3 + 5ab^4 + b^5$

Therefore: $$\left(x^3 + \frac{a}{x^2}\right)^5 = -270$$ $$x^{15}+5x^{10}a+10x^5a^2+10a^3+5\frac{a^4}{x^5}+\frac{a^5}{x^{10}} = -270$$

Which does not help you much, because you do not know the value of $x$, and thus the equation is unsolvable, because it contains two variables. However, it can be solved if $x$ is known. For example, if $x=1$, then you could have done it much easier:

$$5a+10a^2+10a^3+5a^4+a^5 = -271$$

Without knowing the exact value of $x$, this is clealy unsolvable.