
Mark H. answered 12/05/19
Tutoring in Math and Science at all levels
f(x)= x6+16x3+64
Let z = x3 , then the equation becomes:
z2 + 16z + 64
This is (z + 8)*(z + 8)
Replace z with x3 and we have (x3 + 8)*(x3 + 8)
Set either one equal to zero, and we have:
x3 + 8 = 0
x3 = -8
x = -2 (The only zero---AKA "root")


Mark H.
I'm not seeing where you get that....I can divide x^3 + 8 by x + 2 and get x^2 - 2x + 2, which has zeros at 1 +/- i12/05/19
Mark M.
-2 is the only real root. The two imaginary are 1 plus or minus sqrt 312/05/19