If the roots of the equation (b−c)x2+(c−a)x+(a−b)=0 are equal,

The given quadric equation is $(b-c) x^{2}+(c-a) x+(a-b)=0$, and roots are real

Then prove that $2 b=a+c$.

Here,

$a=(b-c), b=(c-a)$ and, $c=(a-b)$

As we know that $D=b^{2}-4 a c$

Putting the value of $a=(b-c), b=(c-a)$ and, $c=(a-b)$

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

$=(c-a)^{2}-4 \times(b-c) \times(a-b)$

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

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

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

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

As we know that $\left(a^{2}+4 b^{2}+c^{2}+2 c a-4 a b-4 b c\right)=(a+c-2 b)^{2}$

$D=(a+c-2 b)^{2}$

The given equation will have real roots, if $D=0$

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

Square root both side we get

$\sqrt{(a+c-2 b)^{2}}=0$

$a+c-2 b=0$

$a+c=2 b$

Hence $2 b=a+c$