If a + b + c = 9 and a2 + b2 + c2 = 35, Find the value of a3 + b3 + c3 - 3abc

Solution:

Given,

$a+b+c=9$ and $a^{2}+b^{2}+c^{2}=35$

We know that,

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

$9^{2}=35+2(a b+b c+c a)$

$81=35+2(a b+b c+c a)$

$81-35=2(a b+b c+c a)$

$46 / 2=a b+b c+c a$

 

$a b+b c+c a=23$

we know that, $a^{3}+b^{3}+c^{3}-3 a b c=(a+b+c)\left(a^{2}+b^{2}+c^{2}-a b-b c-c a\right)$

$a^{3}+b^{3}+c^{3}-3 a b c=(a+b+c)\left[\left(a^{2}+b^{2}+c^{2}\right)-(a b+b c+c a)\right]$ here,

$a+b+c=9$

$a b+b c+c a=23$

 

$a^{2}+b^{2}+c^{2}=35 a^{3}+b^{3}+c^{3}-3 a b c=9[(35-23)]=9 * 12=108$

Hence, the value of $a^{3}+b^{3}+c^{3}-3 a b c$ is 108