Let $a,b,c\in \mathbb{Q}$. I need hint to show that $$a^3+25b^3+5c^3-15abc=0$$ only if $$a=b=c=0.$$
Asked
Active
Viewed 252 times
2
-
https://math.stackexchange.com/questions/475354/how-to-show-that-a3b3c3-3abc-abcab-omegac-omega2ab-omega2 – lab bhattacharjee Nov 08 '18 at 18:13
-
think with $ a^3+ b^3+ c^3 -3abc =0 $ condition. also – maveric Nov 08 '18 at 18:22
-
2With $a=0,b=1,c=-1$ one gets $a^3+5b^3+5c^3-15abc=0.$ – Viera Čerňanová Nov 08 '18 at 19:07
-
Multiplying by their LCM$^3$, we get the same thing with integers. – marty cohen Nov 08 '18 at 19:14
-
2@user376343 Nice counterexample ! – GNUSupporter 8964民主女神 地下教會 Nov 08 '18 at 19:28
-
There was a mistake. I fixed it. I am sorry about it. – Ashot Nov 08 '18 at 20:23
1 Answers
6
Hint: First clear the denomiators and then use infinite decent! So, we can assume that $a,b,c$ are integer.
From $$a^3+25b^3+5c^3-15abc=0 \implies 5\mid a$$
So $a= 5a'$ and thus $$ 25a'^3+5b^3+c^3-15a'bc=0$$
so we get essentialy equation of the same form. But we can 't proceede this inifinite times unless $a=b=c=0$.
nonuser
- 91,557