If a ,b and c are the roots of x^3+ qx+r=0, then (a-b )^2+ (b-c )^2+ (c-a )^2=

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Correct answer: A. (a) Given Equation, x^3+ qx+r=0 Since, a ,b and c are the roots of equation a+b+c=0 ab+ bc+ ca=q … . (i ) and abc=-r As we have, a+b+c=0 (a+b+c )^2=0 a^2+b^2+c^2+2 ab+2 bc+2 ca=0 a^2+b^2+c^2+2( ab+ bc+ ca)=0 a^2+b^2+c^2=-2q Now, (a-b )^2+ (b-c )^2+ (c-a )^2 =a^2+b^2-2 ab+b^2+c^2-2 bc+c^2+a^2-2 ca =2a^2+2b^2+2c^2-2 ( ab+ bc+ ca ) =2(a^2+b^2+c^2)-2 ( ab+ bc+ ca ) [From Eq. (i)] =2(-2q)-2 (q ) =-4q-2q=-6q

$If a, b and c are the roots of x^{3} + qx + r = 0, then (a - b)^{2} + (b - c)^{2} + (c - a)^{2} =$

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