What is the sum of all real values of x that satisfy the equation (x-3)(x^2+4x+4)=x-3+x^3-7x+8?
+1224 (x - 3)(x^2 + 4x + 4) = x - 3 + x^3 - 7x + 8
x^3 + x^2 - 8x - 12 = x^3 - 6x + 5
x^2 - 8x - 12 = -6x + 5
x^2 - 2x - 17 = 0
From there, use the quadratic formula and add the roots to get the answer.
