How would you solve the polynomial equation x^6+7x^3-8=0

I need an explanation too, not just the answer plz

I would substitute u = x³ and solve it as a quadratic in u. Then take the cube roots of u as the solutions for x.

u² + 7u - 8 = 0
(u+8)(u-1) = 0
u = -1, 8

u = -1 = e^(2n+1)πi
x = u^(1/3) = e^(2n+1)πi/3 = e^(iπ/3), e^(πi), e^(5πi/3)
= cosπ/3 + isinπ/3, cosπ + isinπ, cos5π/3 + isin5π/3
= (1+i√3)/2, -1, (1-i√3)/2

u = 8 = 8e^(2nπi)
x = u^(1/3) = 2e^(2nπi/3) = 2, 2e^(2πi/3), 2e^(4πi/3)
= 2, 2(cos2π/3 + isinπ/3), 2(cos4π/3 + isin4π/3)
= 2, -(1+i√3), -(1-i√3)

The solution set contains 2 real and 4 complex roots which are:
x = -1, 2, (1+i√3)/2, (1-i√3)/2, -(1+i√3), -(1-i√3)

add the 8 to the other side, factor the rest.