If a + b + c = 6 and ab + bc + ca = 10, then value of a3 + b3 + c3 - 3abc is :

Answer: Option A

Given ,
a + b + c = 6
ab + bc + ca = 10
∴ (a + b + c)² = 36
⇒ a² + b² + c² + 2ab + 2bc + 2ca = 36
⇒ a² + b² + c² + 2(ab + bc + ca) = 36
⇒ a² + b² + c² + 2 × 10 = 36
⇒ a² + b² + c² = 16
As we know :
$$\Rightarrow \frac{a^3+b^3+c^3-3abc}{a^2+b^2+c^2-ab-bc-ca}$$       $$=\left(a+b+c\right)$$
$$\begin{array}{rl}& \Rightarrow \frac{a^3+b^3+c^3-3abc}{16-\left(ab+bc+ca\right)}=6\\ & \Rightarrow \frac{a^3+b^3+c^3-3abc}{16-10}=6\\ & \Rightarrow a^3+b^3+c^3-3abc=6\times 6\\ & \Rightarrow a^3+b^3+c^3-3abc=36\end{array}$$