Question

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

Answer 1

We have $a+b+c=9$ ............(i)
$\Rightarrow (a+b+c{)}^2=81$ [On squaring both sides of (i)]
$\Rightarrow a^2+b^2+c^2+2(ab+bc+ac)=81$
$\Rightarrow a^2+b^2+c^2+2\times 26=81$ [$\because ab+bc+ac=26$]....(ii)
$\Rightarrow a^2+b^2+c^2=(81-52)\Rightarrow a^2+b^2+c^2=\mathrm{29...}\left(iii\right)$
Now we have
$a^3+b^3+c^3-3abc$
$=(a+b+c)(a^2+b^2+c^2-ab-bc-ac)$
$=(a+b+c)\left[\right(a^2+b^2+c^2)-(ab+bc+ac\left)\right]$ [From (i), (ii), and (iii)]
$=9\times \left[\right(29-26\left)\right]$
$=(9\times 3)=27$

Hence, $a^3+b^3+c^3-3abc=7$