How do I find the value of $k$ in the expansion?

Question

In the expansion of $\left(5x^3 + \dfrac{k}{x^2}\right)^8$ in descending powers of $x$, where $k$ isn’t equal to zero and the coefficient of $x^4$ is equal to the coefficient of $x^9$. How do I find the value of $k$?

Answer 1

Hint:

The $i$th term in the binomial expansion ($0\le i\le 8$) is $$\binom8i (5x^3)^{8-i}\frac{k^i}{x^{2i}}=\binom8i 5^{8-i}k^i x^{24-5i}.$$ Determine the vakues of $i$ which produce the terms in $x^4$ and in $x^9$,and deduce the equation in $k$ which expresses the equality of the coefficients.

Answer 2

Use binomial formula: $$\left(5x^3 + \dfrac{k}{x^2}\right)^8=\sum_{n=0}^8{8\choose n}(5x^3)^n(kx^{-2})^{8-n}=\sum_{n=0}^8 {8\choose n}5^nk^{8-n}x^{5n-16}$$ So: $$5n-16=4 \Rightarrow n=4; \\ 5n-16=9 \Rightarrow n=5$$ Can you finish? Answer:

$${8\choose 4}5^4\cdot k^4={8\choose 5}5^5\cdot k^3 \Rightarrow k=4.$$