If a b+b c+c a=8 and a^{2}+b^{2}+c^{2}=20, then a possible value of \frac{1}{2}(a+b+c) [(a-b)^{2}+(b-c)^{2} +(c-a)^{2} ] is :

SSC CGL 6 06 2019 Shift 2 Paper

Question : 56 of 100

Marks: +1, -0

a b+b c+c a=8

and

a^{2}+b^{2}+c^{2}=20

, then a possible value of

\ {1}/{2}(a+b+c)\ [(a-b)^{2}+(b-c)^{2}

+(c-a)^{2}\ ]

is :

Solution:

(a+b+c)^{2}=20+2 \× 8

=20+16=36

a+b+c=6

find the value
,

\ {1}/{2} \× 6\ [2 a^{2}+2 b^{2}+2 c^{2}

-2(a b+b c+c a)\ ]

=3[2 \× 20-2 \× 8]

=3[40-16]=72

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