Using the sum of cubes formula a^3+b^3 = (a+b)(a^2-ab+b^2)a3+b3=(a+b)(a2−ab+b2) we have:
x^3+4 = x^3+(4^(1/3))^3x3+4=x3+(413)3
=(x+4^(1/3))(x^2-4^(1/3)x+4^(2/3))=(x+413)(x2−413x+423)
If we wish to restrict ourselves to real numbers, we are done. If we allow for complex numbers, then we may continue using the quadratic formula to factor x^2-4^(1/3)x+4^(2/3)x2−413x+423
x = (4^(1/3)+-sqrt(4^(2/3)-4(1)(4^(2/3))))/2x=413±√423−4(1)(423)2
=2^(-1/3)(1+-sqrt(-3))=2−13(1±√−3)
=2^(-1/3)(1+-sqrt(3)i)=2−13(1±√3i)
Thus
x^3+4 = (x+4^(1/3))(x-2^(-1/3)(1+sqrt(3)i))(x-2^(-1/3)(1-sqrt(3)i))x3+4=(x+413)(x−2−13(1+√3i))(x−2−13(1−√3i))
