I think this is trivial

It is what it is

somebodyyouusedtoknow wrote: Problem 2. Let $P(x) = ax^2 + bx + c$ where $a, b, c$ are real numbers. If $$P(a) = bc, \hspace{0.5cm} P(b) = ac, \hspace{0.5cm} P(c) = ab$$then prove that $$(a - b)(b - c)(c - a)(a + b + c) = 0.$$ $p(a)=bc, p(b)=ac, p(c)=ab$ can easily give $-a=ab+ac+b, -b=a^2+ac+b, -c=a^2+ab+b\implies a=b=c$

@Maths_1729 still curious how could you get your last 3 equation

Consider q(x)=x*p(x)-abc. Either 2 of a, b, and c are equal, or a, b, and c are the three distinct roots of q(x)

@QiushengJLi, we know that a, b, c are the root of Q(X) but how do you prove the equation? Still K don't get why can we conclude either 2 of a, b, c are equal Maybe anyone could help for the explanations?

Here's the intended solution (Apparently most participants just straightforwardly bash this problem) Assume that all $a,b,c$ are distinct. We'll prove that $a + b + c = 0$. Let $Q(x) = xP(x) - abc= ax^3 + bx^2 + cx - abc$, then $a,b,c$ are roots of $Q(x)$. Therefore, we get that \[ k(x - a)(x - b)(x - c) = Q(x) = ax^3 + bx^2 + cx - abc \]It's easy to get that $k = 1$ by testing $x =0 $.Thus, we have \[ (x - a)(x - b)(x - c) = ax^3 + bx^2 + cx - abc \]Therefore, it follows that $a = 1$, $2b + c + 1 = 0$ and $b(c + 1) = 0$ by comparing coefficients. We also conclude that $b = 0$ and $c = -1$, which forces $a + b + c = 0$.

Actually I got the same, what I consider since some of the hint above tell they got Either 2 of a, b, and c Which I can't prove and still curious, easily we can bash a+b+c=0 and I knew that it still a right answer, but just want to know it there are prove for the equality for either a, b or b, c or c, a. But nevermind if still there isn't explanation for that part , hopefully I am not disturbing others in the forums

we have $a^3+ab+c=bc$ $ab^2+b^2+c = ac$ $a(a^2-b^2)+ab-b^2=c(b-a)$ if $a = b$ then we are done symetrically for others 3 eqn $(a=b,b=c,c=a)$ thus we just need to check case where $a^3+ab+c-(ab^2+b^2+c)= c(b-a)$ $a(a^2-b^2)+b(a-b)+c(a-b) = 0$ $a(a+b)+b+c = 0$ $a^2+ab+b+c = 0$ $ac+ab+a+b=0$ $a^2+ac+2b = 0$ forcing $a = 1$ and resulting $b+c = -1$ meaning $a+b+c =0$ we are done

TonTony wrote: Actually I got the same, what I consider since some of the hint above tell they got Either 2 of a, b, and c Yep, we just have to prove the fact that a+b+c =0