this problem is from I.M Gelfand's book called Algebra.
Problem 165. Assume that
$\left\{\begin{array}116a+4b+c=0\\49a+7b+c=0\\100a+10b+c=0\end{array}\right.$
Prove that $a=b=c=0$
I know I could simply solve this equation for $a,b,c$ and show that they're equal to 0 but I wonder if there's more smart way of showing it, since this system looks like:
$\left\{\begin{array}(x_1^2a+x_1b+c=0\\x_2^2a+x_2b+c=0\\x_3^2a+x_3b+c=0\end{array}\right.$