Question
If $a+b+c=9$ and $a^2+b^2+c^2=35$, find value of $a^3+b^3+c^3-3 a b c$
✓
Answer
We know that
$a^3+b^3+c^3-3 a b c=(a+b+c)\left(a^2+b^2+c^2-a b-b c-c a\right)$
$\Rightarrow a^3+b^3+c^3-3 a b c=(a+b+c)\left[\left(a^2+b^2+c^2\right)-(a b+b c+c a)\right] \ldots(1)$
It follow from the above identity that we require the values of $a+b+c, a^2+b^2+c^2$, and $a b+b c+c a$ to get the value of $a^3+b^3+c^3-3 a b c$.
The values of $a+b+c$ and $a^2+b^2+c^2$ are known to us.
So we require the value of $a b+b c+c a$,
Now,
$(a+b+c)^2=a^2+b^2+c^2+2(a b+b c+c a)$
$\Rightarrow(9)^2=35+2(a b+b c+c a)\left[\therefore a+b+c=9 a n d a^2+b^2+c^2=35\right]$
$\Rightarrow 81=35+2(a b+b c+c a)$
$\Rightarrow 2(a b+b c+c a) 81-35=46$
$\Rightarrow a b+b c+c a=\frac{46}{2}=23$
Substituting the values of $a b+b c+c a$ in (1), we get,
$a^3+b^3+c^3-3 a b c=9(35-23)\left(\therefore a+b+c=9 a n d a^2+b^2+c^2=35\right)$
$=9 \times 12$
$=108$
$\therefore a^3+b^3+c^3-3 a b c=108$
$a^3+b^3+c^3-3 a b c=(a+b+c)\left(a^2+b^2+c^2-a b-b c-c a\right)$
$\Rightarrow a^3+b^3+c^3-3 a b c=(a+b+c)\left[\left(a^2+b^2+c^2\right)-(a b+b c+c a)\right] \ldots(1)$
It follow from the above identity that we require the values of $a+b+c, a^2+b^2+c^2$, and $a b+b c+c a$ to get the value of $a^3+b^3+c^3-3 a b c$.
The values of $a+b+c$ and $a^2+b^2+c^2$ are known to us.
So we require the value of $a b+b c+c a$,
Now,
$(a+b+c)^2=a^2+b^2+c^2+2(a b+b c+c a)$
$\Rightarrow(9)^2=35+2(a b+b c+c a)\left[\therefore a+b+c=9 a n d a^2+b^2+c^2=35\right]$
$\Rightarrow 81=35+2(a b+b c+c a)$
$\Rightarrow 2(a b+b c+c a) 81-35=46$
$\Rightarrow a b+b c+c a=\frac{46}{2}=23$
Substituting the values of $a b+b c+c a$ in (1), we get,
$a^3+b^3+c^3-3 a b c=9(35-23)\left(\therefore a+b+c=9 a n d a^2+b^2+c^2=35\right)$
$=9 \times 12$
$=108$
$\therefore a^3+b^3+c^3-3 a b c=108$
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