how do you factor x^3 - 1?

Question

does anyone know how to factor 15cubedycubed + 10xsquaredysquared + 5xy????

Answer 1

OK

This is a perfect cube

x^3 -1 =
(x-1)(x^2+x+1)

I think you may have missed an x is you second statement so I will go with this:


15x^3y^3 + 10x^2y^2 + 5xy

5xy(3x^2y^2 +2xy+1)

Hope that helps.

Answer 2

15x³y³ + 10x²y² + 5xy = 0

Divide both sides by 15xy:
x²y² + (2/3)xy + (1/3) = 0

Complete the square:
x²y² + (2/3)xy = -(1/3)
x²y² + (2/3)xy + (1/3)² = -(1/3) + (1/3)²
(xy + 1/3)² = -3/9 + 1/9
(xy + 1/3) = ± √(-2/9)
xy = -1/3 ± (1/3)i√2
xy₊ = -1/3 + (1/3)i√2
xy₋ = -1/3 - (1/3)i√2

Answer 3

a^3 – b^3 = (a – b)(a^2 + ab + b^2)

so

x^3 - 1 = (x-1)(x^2 + x + 1)

to do
I'm going to assume you left out x after the 15
15 x^3 y^3 + 10x^2 y^2 + 5x y
factor out the monomial 5xy
5xy (3x^2 y^2 + 2x y + 1)
this is finished because the discriminant of the trinomial is -8

Answer 4

(x - 1)(x²+x+1)

Answer 5

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

Answer 6

I don't think (x-1)(x^2 + x +1) is right. I think that gives
X^3 + 2x^2 - 1. Rather, just
(x^2 + 1) (x - 1)

Answer 7

(X^3 -1 ) = (x - 1)(x²+x+1)

(15x^3y^3 + 10x²y² + 5xy )
=xy(15x²y² + 10xy + 5 )