Question

If the roots of the equation
$(c^2-ab)x^2-2(a^2-bc)x+(b^2-ac)=0$
be equal, prove that either $a=0$
or $a^3+b^3+c^3+=3abc$.

Answer 1

$(c^2-ab)x^2-2(a^2-bc)x+(b^2-ac)=0$
If the roots be equal, then $B^2-4AC=0$
$\therefore$ $4{(a^2-bc)}^2-4(c^2-ab)(b^2-ac)=0$
or $[a^4-2a^{2}bc+b^2{c}^2]-[b^2{c}^2-a{b}^3-a{c}^3+a^{2}bc]=0$
or $a[a^3+b^3+c^3+-3abc]=0$
$\therefore$ Either $a-0$ or $a^3+b^3+c^3-3abc=0$.