Question

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

Answer 1

We have,

$a+b+c=9$

$ab+bc+ca=26$

We know that

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

$a^3+b^3+c^3-3abc=(a+b+c)\left(\right(a+b+c{)}^2-2(ab+bc+ca)-(ab+bc+ca))$

$a^3+b^3+c^3-3abc=(a+b+c)\left(\right(a+b+c{)}^2-3(ab+bc+ca))$

Therefore,

$a^3+b^3+c^3-3abc=\left(9\right)\left(\right(9{)}^2-3\left(26\right))$

$a^3+b^3+c^3-3abc=\left(9\right)(81-78)$

$a^3+b^3+c^3-3abc=\left(9\right)\left(3\right)=27$

Hence, this is the answer.