x^2(3x^2+k/x)^8 The constant is 16128. Find k

Question

 This is binomial expansion. I need to find k. The constant is 16128. Can you please help me?

 

x²(3x²+k/x)^8.  The constant is 16128. Find k

 

Answer 1

I am assuming that you mean that the constant TERM is 16128.

 

We need a term in the expansion of (3x² + k/x)⁸ such that the power of x is 1/x² (to cancel out the x² in front of the parentheses).

 

The 7th term is ₈C₆(3x²)²(k/x)⁶ = 252k/x²

 

So, 252k = 16128

 

           k = 64