(x + 3)(x^2 - 3x + 9)
t3-1/27
w3+125
x3 + x = x(x2+1)
2(x^2 + 2)(x + 3)
X^3 + 27 = 0 subtract 27 from both sides X^3 = -27 take cubic root of both sides X = -3 (this sentence is true for x = -3) x3 + 27 = 0 the left side is the sum of the cubes, so it is factorable such as: x3 + 33 = 0 (x + 3)(x2 - 3x + 9) = 0
(x + y)(x + y)(x + y)
t3-1/27
x(x^2 + 1)
x(x2 + 36)
(x - 3)(x2 + 3x + 9)
w3+125
Factor out the GCF and get X(X2-X+1).
x(x + 9)(x + 1)
x(x + 9)(x + 1)
Since the problem has 4 terms, first you factor x cubed plus 9x squared, then you factor 2x plus 18. So when you factor the first two term, you would get x sqaured (x plus 9). Then when you factor the last two terms and you get 2 (x plus 9). Ypure final answer would be (x squared plus 2)(x plus 9)
x^3 + y^3 = (x + y)(x^2 - xy + y^2)
(x + 6)(x^2 - 6x + 36)