If the roots of the equation

Given:

$\left(c^{2}-a b\right) x^{2}-2\left(a^{2}-b c\right) x+\left(b^{2}-a c\right)=0$

Here,

$a=\left(c^{2}-a b\right), b=-2\left(a^{2}-b c\right), c=\left(b^{2}-a c\right)$

It is given that the roots of the equation are real and equal; therefore, we have:

$D=0$

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

$\Rightarrow\left\{-2\left(a^{2}-b c\right)\right\}^{2}-4 \times\left(c^{2}-a b\right) \times\left(b^{2}-a c\right)=0$

$\Rightarrow 4\left(a^{4}-2 a^{2} b c+b^{2} c^{2}\right)-4\left(b^{2} c^{2}-a c^{3}-a b^{3}+a^{2} b c\right)=0$

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

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

$\Rightarrow a\left(a^{3}-3 a b c+c^{3}+b^{3}\right)=0$

Now,

$a=0$ or $a^{3}-3 a b c+c^{3}+b^{3}=0$

$a=0$ or $\mathrm{a}^{3}+b^{3}+c^{3}=3 a b c$