If a + b + c = 15 and a2 + b2 + c2 = 83 then the value of a3 + b3 + c3 - 3abc = ?

Answer: Option B

$$\begin{array}{rl}& a+b+c=15\\ & a^2+b^2+c^2=83\left(\text{Given}\right)\\ & \therefore a+b+c=15\\ & \left(\text{Squaring both sides}\right)\\ & \Rightarrow {\left(a+b+c\right)}^2={\left(15\right)}^2\\ & \Rightarrow a^2+b^2+c^2+2ab+2bc+2ca=225\\ & \Rightarrow 83+2\left(ab+bc+ca\right)=225\\ & \Rightarrow 2\left(ab+bc+ca\right)=142\\ & \Rightarrow ab+bc+ca=71\end{array}$$
  $$\therefore a^3+b^3+c^3-3abc=$$     $$\left(a+b+c\right)$$  $$\left(a^2+b^2+c^2-ab-bc-ca\right)$$
$$\begin{array}{rl}& \Rightarrow a^3+b^3+c^3-3abc=15\left(83-71\right)\\ & \Rightarrow a^3+b^3+c^3-3abc=15\times 12\\ & \Rightarrow a^3+b^3+c^3-3abc=180\end{array}$$