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

$a+b+c=9$
Squaring on both sides
$(a+b+c{)}^2=81$
$a^2+b^2+c^2+2ab+2bc+2ca=81$
$a^2+b^2+c^2+2(ab+bc+ca)=81$
$a^2+b^2+c^2+2\left(26\right)=81$
$a^2+b^2+c^2=81–52=29$
Now, $a^3+b^3+c^3-3abc$
$=(a+b+c)(a^2+b^2+c^2–ab–bc-ca)$
$=(a+b+c)\left(\right(a^2+b^2+c^2)-(ab+bc+ca\left)\right)$
$=\left(9\right)(29–26)$
$=9\times 3=27$