If $a +b + c = 9$ and $ab + bc + ca = 26$, find the value of $a^3 + b^3 + c^3 - 3abc$.

Given :

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

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,

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

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

$a^2 + b^2 + c^2 + 2 \times26 = 81$

$a^2 + b^2 + c^2 + 52 = 81$

$a^2 + b^2 + c^2 = 81 - 52 = 29$

Therefore,

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

$= 9(29 – 26)$

$= 9 \times 3$

$= 27$

Hence, $a^3 + b^3 + c^3 – 3abc =27$.