Math
posted by Mike on .
(Factoring a sum or difference of cubes) and (solving a polynomial equation).
2x^3+2=0

2x^3+2=0
divide by 2
x^3 +1=0
now you have a sum of cubes
(x+1)(x^2x+1)=0
x=1 or x = (1 ± ?3)/2 or (1 ± i?3)/2 
Factor the given equation into:
2x³+2 =0
=> 2(x²+1)=0
and use one of the identities:
x³+y²=(x+y)(x²xy+y²)
x³y²=(xy)(x²+xy+y²)
to factor the given equation to linear and quadratic factors.
Hence solve the resulting linear and quadratic equations.