Question

If $c^2\ne ab$, and the root of$(c^2-ab)x^2-2(a^2-bc)x+(b^2-ac)=0$ are equal then show that $a^3+b^3+c^3=3abc$ or a =0

Answer 1

$(c^2-ab)x^2-2(a^2-bc)x+(b^2-ac)=0$

It is of form $A{x}^2+Bx+C=0$

For the roots to be equal

$\Rightarrow B^2=4AC$

$4(a^2-bc{)}^2=4(c^2-ab\left)\right(b^2-ac)$

$\Rightarrow a^5+b^2{c}^2-2a^{2}bc=b^2{c}^2-a{c}^3-a{b}^3+a^{2}bc$

$\Rightarrow a^4+a{c}^3+a{b}^3=3a^{2}bc$

$\Rightarrow a(a^3+b^3+c^3)=a\left(3abc\right)$

$\Rightarrow Eithera=0$ or if $a\ne 0,a^3+b^3+c^3=3abc$