If the quadratic equation $(b^2+c^2)x^2-2c(a+b)x+(c^2+a^2)=0$ has equal roots then, $a,c,b$ are in

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

$a=b^2+c^2,b=-2c(a+b),c=c^2+a^2$

$b^2-4ac=0$

$b^2=4ac$

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

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

$c^2(a^2+b^2+2ab)=b^2{c}^2+b^2{a}^2+c^4+a^2{c}^2$

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

$2ab{c}^2-b^2{a}^2-c^4=0$

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

${\left(c^2\right)}^2+{\left(ba\right)}^2-2\times ab\times c^2=0$

${(c^2-ab)}^2=0$

$c^2-ab=0$

$c^2=ab$

$a,c,b$ are in A.P