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

Answer 1

k = 2 or -2

3rd term in the binomial expansion has a term with x^-2. multiply it by x^2 and you get the only constant term in the product.

8C2(k/x)^6(3x^2)^2 = 256k^6/x^2

times x^2 = 256k^6 = 16128

k^6 = 16128/256 = 64

k = 6th root of 64. The only real roots are 2 and -2

there are 4 other imaginary values for k

k^6-64 = 0

6th degree equation has 6 solutions