If a + b + c = 9 and ab + bc + ca = 26, Find the value of

Solution:

Given,

a + b + c = 9 and ab + bc + ca = 26

We know that,

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

$(a+b+c)^{2}=a^{2}+b^{2}+c^{2}+2(26)$

$(a+b+c)^{2}=a^{2}+b^{2}+c^{2}+52$

$9^{2}=a^{2}+b^{2}+c^{2}+52$

$81-52=a^{2}+b^{2}+c^{2}$

$a^{2}+b^{2}+c^{2}=29$

we know that,

$a^{3}+b^{3}+c^{3}-3 a b c=(a+b+c)$

$\left(a^{2}+b^{2}+c^{2}-a b-b c-c a\right) a^{3}+b^{3}+c^{3}-3 a b c=(a+b+c)\left[\left(a^{2}+b^{2}+c^{2}\right)-(a b+b c+c a)\right] h e r e$

$a+b+c=9$

$a b+b c+c a=26$

$a^{2}+b^{2}+c^{2}=29$

$a^{3}+b^{3}+c^{3}-3 a b c=9[(29-26)]=9 * 3=27$

Hence, the value of $a^{3}+b^{3}+c^{3}-3 a b c$ is 27