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

Answer 1

Compare given equation with the general form of quadratic equation, which $a{x}^2+bx+c=0$

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

Since roots are equal, so $D=0$

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

$4(a^4-2a^{2}bc+b^2{c}^2)-4(b^2{c}^2-a{c}^3-ab3+a^{2}bcz)=0$

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

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

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

$a=0$ or $a^3+b^3=3abc$

Hence proved.