Question

a+b+c=0,a²+b²+c²/b²-c²=?​

Answer 1

Given,

$\:\frac{a^2+b^2+c^2}{b^2-c^2}=0\\a=\sqrt{-b^2-c^2}\\a=-\sqrt{-b^2-c^2}\\Using\:the\:Zero\:Factor\:Principle\\ab=0\:\mathrm{then}\:a=0\:\mathrm{or}\:b=0\\a^2+b^2+c^2=0\\\mathrm{Subtract\:}b^2+c^2\mathrm{\:from\:both\:sides}\\a^2+b^2+c^2-\left(b^2+c^2\right)=0-\left(b^2+c^2\right)\\a^2=-\left(b^2+c^2\right)\\a^2=-b^2-c^2\\\mathrm{For\:}x^2=f\left(a\right)\mathrm{\:the\:solutions\:are\:}x=\sqrt{f\left(a\right)},\:\:-\sqrt{f\left(a\right)}\\a=\sqrt{-b^2-c^2}\\a=-\sqrt{-b^2-c^2}$