If $a + b + c = 9$, and $a^2 + b^2 + c^2 = 35$, find the value of $a^3 + b^3 + c^3 - 3abc$.

Given :

$a + b + c = 9$, and $a^2 + b^2 + c^2 = 35$

To do :

We have to find the value of $a^3 + b^3 + c^3 - 3abc$.

Solution :

We know that,

$(a+b+c)(a^2+b^2+c^2-ab-bc-ca)=a^3+b^3+c^3-3abc$

$a + b + c = 9$

Squaring both sides, we get,

$(a + b + c)^2 = (9)^2$

$a^2 + b^2 + c^2 + 2 (ab + bc + ca) = 81$

$35 + 2 (ab + bc + ca) = 81$

$2(ab + bc + ca) = 81-35$

$(ab + bc + ca) = \frac{46}{2} = 23$

Therefore,

$a^3 + b^3 + c^3 – 3abc = (a + b + c) [(a^2 + b^2 + c^2) – (ab + bc + ca)]$

$= 9(35 – 23)$

$= 9 \times 12$

$= 108$

Hence, $a^3 + b^3 + c^3 – 3abc =108$.