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, $a^3 + b^3 + c^3 – 3abc =108$.
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