### Reformatting the input :

Changes made to your input must not influence the solution: (1): "x3" was reput by "x^3".

You are watching: How to factor x^3-8

## Tip 1 :

Trying to variable as a Difference of Cubes:1.1 Factoring: x3-8 Theory : A difference of 2 perfect cubes, a3-b3 have the right to be factored into(a-b)•(a2+ab+b2)Proof:(a-b)•(a2+ab+b2)=a3+a2b+ab2-ba2-b2a-b3 =a3+(a2b-ba2)+(ab2-b2a)-b3=a3+0+0-b3=a3-b3Check:8is the cube of 2Check: x3 is the cube of x1Factorization is :(x - 2)•(x2 + 2x + 4)

Trying to variable by separating the middle term

1.2Factoring x2 + 2x + 4 The initially term is, x2 its coeffective is 1.The middle term is, +2x its coreliable is 2.The last term, "the constant", is +4Step-1 : Multiply the coreliable of the first term by the continuous 1•4=4Step-2 : Find 2 determinants of 4 whose sum amounts to the coefficient of the middle term, which is 2.

 -4 + -1 = -5 -2 + -2 = -4 -1 + -4 = -5 1 + 4 = 5 2 + 2 = 4 4 + 1 = 5

Observation : No 2 such determinants have the right to be found !! Conclusion : Trinomial have the right to not be factored

## Final outcome :

(x - 2) • (x2 + 2x + 4)

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