Find all triplets of real numbers \((a,b,c)\) with \(a^2+b^2=c^2\) satisfying

\[2(a^3+b^3+c^3)=ab(3a+3b-4c)+bc(3b+3c-4a)+ca(3c+3a-4b)\]

Find all triplets of real numbers \((a,b,c)\) with \(a^2+b^2=c^2\) satisfying

\[2(a^3+b^3+c^3)=ab(3a+3b-4c)+bc(3b+3c-4a)+ca(3c+3a-4b)\]

No vote yet

1 vote

×

Problem Loading...

Note Loading...

Set Loading...

## Comments

Sort by:

TopNewestHow could we rewrite this using cyclic sums? – Cody Johnson · 3 years, 1 month ago

Log in to reply

We can assume \(a=c\sin\theta,b=c\cos\theta\) then solve: \[3(\sin^3\theta+\cos^3\theta+1)+18(\sin\theta\cos\theta)=(\sin\theta+\cos\theta+1)^3,\] that leads to: \[ \cos(\theta/2)\cdot\sin(\pi/4+\theta/2)\cdot (\cos(\theta/2)-3\sin(\theta/2))\cdot(\cos(\theta/2)-2\sin(\theta/2))\cdot (\sin \theta+\cos \theta-2)=0,\] hence \(\theta\) can take four values in \((-\pi,\pi]\): \(-\pi/2,2\arctan(1/3),2\arctan(1/2),\pi\). – Jack D'Aurizio · 3 years, 1 month ago

Log in to reply

– Cody Johnson · 3 years, 1 month ago

So what are our possible values of \((a,b,c)\)?Log in to reply