AP EAMCET Engineering 2018 Apr 22 Shift 1 Paper

The equation is given as.
x3+qx+r=0
The roots of the given equation is.
a+b+c=0
ab+bc+ca=q
abc=−r
Squaring the root of the equation.
(a+b+c)2=0
a2+b2+c2+2ab+2bc+2ca=0
a2+b2+c2+2(ab+bc+ca)=0
Substituting the values we get,
a2+b2+c2=−2q
Now,
(a−b)2+(b−c)2+(c−a)2
=a2+b2−2ab+b2+c2−2bc +c2+a2−2ac
=2(a2+b2+c2)−2(ab+bc+ca)
=9(−2q)−2(q)
=−6q