If a+b+c=11 and a b+b c+c a=38, then a^{3}+b^{3}+c^{3} -3 a b c is equal to:

SSC CGL 10 Jun 2019 Shift 2 Paper

Question : 68 of 100

Marks: +1, -0

a+b+c=11

and

a b+b c+c a=38,

then

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

-3 a b c

is equal to:

Solution:

(a+b+c)^{2}=11^{2}

\⇒ a^{2}+b^{2}+c^{2}=121-38 \× 2=45

\∴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\ )

=11 \×(45-38)=77

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