Hi, there's this simultaneous equation problem, however I already know the solution, I just want to know more about the solution. a+b+c=2,bc+ac+ab=-5, abc=-6 The solution is: Consider the polynomial (x-a)(x-b)(x-c), where a,b,c are constants. (x-a)(x-b)(x-c) = x^3-(a+b+c)x^2+(bc+ac+ab)x-abc Substitute the given equations into the polynomial: (x-a)(x-b)(x-c) = x^3-2x^2-5x+6 (x-a)(x-b)(x-c) = (x-1)(x+2)(x-3) Therefore, a,b,c = 1,-2,3 in any order. My question is, how can you know that the three equations relate to a cubic purely from inspection? And is this some broader strategy to solve non-linear simultaneous equations? If the normal method of substituting is used to solve these simultaneous equations, a=2-b-c sub a in both equations: b(2-b-c)+c(2-b-c)+bc=-5 b^2+bc+c^2-2b-2c=5 and (2-b-c)bc=-6 bc(b+c-2)=6 b^2c+bc^2-2bc=6 There isn't much else you can do, unless get b+c-2 as the subject of both equations, then sub again, which doesn't yield anything, though it's not surprising that you can't solve this by normal methods, as if you could it would mean you could factorise any polynomial of degree 3, and by extension, of any degree easily, which you can't. Also, just investigating, if you do a similar thing for a quadratic: a+b=-3 ab=2 Again we can see see these are the coefficients of a polynomial, namely a quadratic: (x-a)(x-b) = x^2-(a+b)x+ab sub the given equations in: =x^2+3x+2 =(x+1)(x+2) so a, b=-1,-2 in any order However, we can also solve this through normal methods: a=2/b 2/b+b=-3 2+b^2=-3b b^2+3b+2=0 (b+1)(b+2)=0 b=-1,-2, then solve for a This could suggest you could do something similar for the cubic, however I can't see how.