Question

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

Answer 1

If the roots of the equation $(c^2-ab)x^2-2(a^2-bc)x+b^2-ac=0$ are equal,

then show that either a=0 or $a^3+b^3+c^3=3abc$

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

$a=c^2-ab,b=-2(a^2-bc),c=b^2-ac$

since roots are equal,

$\Rightarrow b^2-4ac=0$

$\therefore b^2=4ac$

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

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

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

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

$a^4-3a^{2}bc+a{c}^3+a{b}^3=0$

$a(a^3+c^3+b^3-3abc)=0$

$\therefore a=0$,

$a^3+c^3+b^3-3abc=0\Rightarrow a^3+c^3+b^3=3abc$