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TIFR-2011-Maths-A-8

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Let $a$,$b$,and $c$ are roots of equations.

So, $(x-a)(x-b)(x-c)=0$

$=(x-a)(x^2-bx-cx+bc)$

$= x^3-ax^2-bx^2-cx^2+abx+bcx+cax-abc$

$=x^3-(a+b+c)x^2+(ab+bc+ca)x-abc $

Compare it with given equation$x^3-4x^2+6x+1$

$a+b+c =4$

$ab+bc+ca=6$

$abc=-1$

now, as we know $(a+b+c)^2= a^2+b^2+c^2+2ab+2bc+2ca$

so, $4^2= a^2+b^2+c^2+2\times6$

So, Sum of squares of roots , $a^2+b^2+c^2=16-12=4$
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