If you re-write P(x) = X^6 + 16X³ + 64 as P(x) = (X³)² + 2*8X³ + 8² you can see that P(X) = (X³ + 8)²

This way you find that there is only 1 (real) zero X = -2 with multiplicity 6 (since the sum of multiplicities must be equal to the degree of the polynomial).

Let me know if you have any further questions.

Best regards,

Tom

ExcelSpecialist and other Math Homework Specialists are ready to help you

Ok now we have (X^3+8)^2 How can you find the zeros of (X^3+8) I have only reduced to a linear factor and a irreducible quadratic. Never saw this type of problem with a X^3 before.