Question 490715
Expanding your equation yields


*[tex \LARGE x^3 + (x^3 + 3x^2 + 3x + 1) + (x^3 + 6x^2 + 12x + 8) = x^3 + 9x^2 + 27x + 27]


*[tex \LARGE 3x^3 + 9x^2 + 15x + 9 = x^3 + 9x^2 + 27x + 27]


*[tex \LARGE 2x^3 - 12x - 18 = 0 \Rightarrow x^3 - 6x - 9 = 0]


From here, we can tell that x = 3 works. If you divide both sides of the previous equation by x-3 (either by synthetic or long division) we can find the other two roots:


*[tex \LARGE x^2 + 3x + 3 = 0]


The discriminant is negative, so there are no more integer roots. Hence x=3 --> {3,4,5,6} is the only solution.