Question:
If $a+b+c=9$ and $a^{2}+b^{2}+c^{2}=35$, find the value of $\left(a^{3}+b^{3}+c^{3}-3 a b c\right)$.
Solution:
$a+b+c=9$
$\Rightarrow(a+b+c)^{2}=9^{2}=81$
$\Rightarrow a^{2}+b^{2}+c^{2}+2(a b+b c+c a)=81$
$\Rightarrow 35+2(a b+b c+c a)=81$
$\Rightarrow(a b+b c+c a)=23$
We know,
$\left(a^{3}+b^{3}+c^{3}-3 a b c\right)=(a+b+c)\left(a^{2}+b^{2}+c^{2}-a b-b c-c a\right)$
$=(9)(35-23)$
$=108$
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