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?\
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 $\displaystyle a^3- a= a(a^2- 1)= a(a-1)(a+1)= 0$.