If a+b+c=8 and a b+b c+c a=12, then a^{3}+b^{3}+c^{3}- 3abc is equal to:

SSC CGL 11 Jun 2019 Shift 1 Paper

Question : 65 of 100

Marks: +1, -0

a+b+c=8

and

a b+b c+c a=12,

then

a^{3}+b^{3}+c^{3}-

3abc is equal to:

Solution:

(a+b+c)^{2}=(8)^{2}

\⇒a^{2}+b^{2}+c^{2}=64-2 \×12

=40

\∵a^{3}+b^{3}+c^{3}-3 a b c

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

-b c-c a\ )

=8 \×(40-12)=224

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