(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)
Factor out the GCF and get X(X2-X+1).
w3+125
x(x + 9)(x + 1)
x(x + 9)(x + 1)
x^3 + y^3 = (x + y)(x^2 - xy + y^2)
(x + 6)(x^2 - 6x + 36)
x^3 + x^2 + x = x(x^2 + x + 1)