Question

If a + b + c = 9 and ab + bc + ca = 23, then $a^3+b^3+c^3-3abc=$

• A. 108
• B. 207
• C. 669
• D. 729

Answer 1

The correct option is A

108

$a^3+b^3+c^3-3abc=(a+b+c)[a^2+b^2+c^2-(ab+bc+ca\left)\right]\text{Now,}a+b+c=9\text{Squaring,}{a}^2+b^2+c^2+2(ab+bc+ca)=81\Rightarrow a^2+b^2+c^2+2\times 23=81\Rightarrow a^2+b^2+c^2+46=81\Rightarrow a^2+b^2+c^2=81-46=35\text{Now},a^3+b^3+c^3-3abc=(a+b+c)\left[\right(a^2+b^2+c^2)-(ab+bc+ca\left)\right]=9[35-23]=9\times 12=108$