A problem where I have tried to use binomial theorem but getting a wrong answer which is $x^5$/42 please help me

Question

If constant term in the expansion of ($x^3$ + $\frac{k}{x^8})^{11}$. Is 1320. Find k. As to my work I found the middle term as it will be a constant one but answer is coming absurd

Answer 1

The constant term will come from the value of $n$, $0 \leq n \leq 11$, such that $$3n = 8(11-n).$$ That is, $n = 8$. The binomial coefficient for the term $$(x^{3})^8 \cdot \left(\frac{k}{x^8}\right)^{3} = k^3$$ is $$\frac{11!}{8!3!} = 165.$$ Given that the constant term equals $1320$, we have $165k^3 = 1320$, that is, $k^3 = 8$. Hence, $k=2$.