Question

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

• A. $27$
• B. $29$
• C. $495$
• D. $729$

Answer 1

The correct option is A $27$

$a+b+c=9$ ...Given

Squaring on both sides

$(a=b+c{)}^2=81$

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

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

$a^2+b^2+c^2+2\left(26\right)=81$

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

$a^3+b^3+c^3-3abc$

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

$=(a+b+c)\left[\right(a^2+b^2+c^2)-(ab+bcca\left)\right]$

$=\left(9\right)(29-26)$

$=9\times 3=27$