Question 557105
*[tex \LARGE x^3 - y^3 = (x-y)(x^2 + xy + y^2)]


x-y is obviously irreducible. For sake of completeness, we show that x^2 + xy + y^2 is irreducible. Suppose that, on the contrary, it is reducible into two polynomials. Then we can write it in a general form:


*[tex \LARGE x^2 + xy + y^2 = (x+ay+b)(x+cy+d)]


Hence by equating coefficients we have ac = 1, ad + bc = 0, a+c = 0, bd = 0. Since bd = 0, one of b or d must be 0. Assume that b = 0 (we can do this due to symmetry). Then ad = 0. Since ac = 1, a cannot equal 0 so d = 0. Now we have ac = 1 and a+c = 0. This does not reveal any solutions in real numbers (since a and c are additive inverses, product is negative) so we have a contradiction and x^2 + xy + y^2 is irreducible.