Question

If the roots of the equation $(c^2-ab)x^2-2(a^2-bc)x+(b^2-ac)=0$ are real and equal, show that either $a=0$ or $a^3+b^3+c^3=3abc$
$[Hint:$ $D=4a(a^3+b^3+c^3-3abc)]$

Answer 1

$(c^2-ab)x^2-2(a^2-bc)x+b^2-ac=0$ has roots real and equal thus its discriminant is zero.

$D=b^2-4a^{\prime }{c}^{\prime }=0$

thus $(-2(a^2-bc){)}^2-4(c^2-ab\left)\right(b^2-ac)=0$

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

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

$=4a(a^3+b^3+c^3-3abc)=0$

$\Rightarrow$ Either $a=0$ or $a^3+b^3+c^3=3abc$.