Posted by **Mike** on Monday, January 31, 2011 at 6:40pm.

(Factoring a sum or difference of cubes) and (solving a polynomial equation).

2x^3+2=0

- Math -
**Reiny**, Monday, January 31, 2011 at 7:00pm
2x^3+2=0

divide by 2

x^3 +1=0

now you have a sum of cubes

(x+1)(x^2-x+1)=0

x=-1 or x = (1 ± ?-3)/2 or (1 ± i?3)/2

- Math -
**MathMate**, Monday, January 31, 2011 at 7:00pm
Factor the given equation into:

2x³+2 =0

=> 2(x²+1)=0

and use one of the identities:

x³+y²=(x+y)(x²-xy+y²)

x³-y²=(x-y)(x²+xy+y²)

to factor the given equation to linear and quadratic factors.

Hence solve the resulting linear and quadratic equations.

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