Prove that, if 1 + 3x^2 + x^3 < (1+x)^3, then x>0

(1+x)^3 = x^3 + 3x^2 + 3x + 1 (Can be calculated straight away by binomial method or by multiplying brackets individually)
if (1+x)^3 > 1 + 3x^2 + x^3then: x^3 + 3x^2 + 3x + 1 > 1 + 3x^2 + x^3 3x > 0 x > 0