# Thread: solve system of equations

1. ## solve system of equations

Solve the system in real $\displaystyle x,y,z:$
$\displaystyle (a+b)^3=8c , (a+c)^3=8b , (b+c)^3=8a$
The solution is clearly $\displaystyle (a,b,c)=(0,0,0),(1,1,1),(-1,-1,-1)$, but I need some trick to "arrive" at it... any ideas?\

2. ## Re: solve system of equations

Originally Posted by atreyyu
Solve the system in real $\displaystyle x,y,z:$
$\displaystyle (a+b)^3=8c , (a+c)^3=8b , (b+c)^3=8a$
The solution is clearly $\displaystyle (a,b,c)=(0,0,0),(1,1,1),(-1,-1,-1)$, but I need some trick to "arrive" at it... any ideas?\
No, you do not need "tricks", you need understanding. The first thing you should see is that these three equations are "symmetric"- if you interchange a, b, or c in any one, you get another one. That tells you that a= b= c. With b and c replaced by a, all three equations reduce to $\displaystyle (a+ a)^3= 8a^3= 8a$ so that $\displaystyle a^3= a$ and $\displaystyle a^3- a= a(a^2- 1)= a(a-1)(a+1)= 0$.

3. ## Re: solve system of equations

Originally Posted by HallsofIvy
That tells you that a= b= c.
I was wondering if we were allowed to make that conjecture...