1. The problem statement, all variables and given/known data Prove that if a² + ab + b² = 0 then a = 0 and b = 0 Hint: Recall the factorization of a³-b³. (Another solution will be discussed later when speaking about quadratic equations.) 2. Relevant equations a² + ab + b² is close to a² + 2ab + b² = (a+b)² a³-b³=(a-b)(a²+ab+b²) 3. The attempt at a solution a² + ab + b² = 0 a² + ab + b² + ab = ab a² + 2ab + b² = ab (a+b)² = ab (a+b)² - ab = 0 Here I figured, to get zero after a subtraction, both terms need to be equal. An addition of two numbers is the same to the multiplication of those two numbers only when a,b = 0 or a,b = 2. (0+0=0*0 and 2+2=2*2) But since the first term is squared, (2+2)² - 2*2 = 16 - 4 ≠0. This leaves a,b = 0 as the only option to get a zero. A second try was a² + ab + b² = 0 a² + ab + b² + ab = ab a² + 2ab + b² = ab (a+b)² = ab (a+b)² - ab = 0 (b+b)² - bb = 0 (given that a=b, and same works for b=a) (2b)² - b² = 0 (2b-b)(2b+b) = 0 b*b = 0 b = 0 A third unsuccessful attempt a² + ab + b² = 0 a² + ab + b² + ab = ab a² + 2ab + b² = ab (a+b)² = ab a+b = √(ab) a+b = √a√b (a+b) - (√a√b) = 0 The solution: a³-b³=(a-b)(a²+ab+b²), if (a²+ab+b²)=0, then a³-b³=0 or a³ = b³ which can only happen when a=b. If a=b, then 0=a²+ab+b²=3b² which implies that b=0 and a=b=0. I have no experience with proofs whatsoever. I understand the solution but I want to know if any of my first three tries have any correctness in them. Or is the solution above the only possible way to prove this equality without any other 'techniques'?