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

Solution(By Examveda Team)

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 :
$$ \Rightarrow \frac{{{a^3} + {b^3} + {c^3} - 3abc}}{{{a^2} + {b^2} + {c^2} - ab - bc - ca}}$$       $$ = \left( {a + b + c} \right)$$
$$\eqalign{ & \Rightarrow \frac{{{a^3} + {b^3} + {c^3} - 3abc}}{{16 - \left( {ab + bc + ca} \right)}} = 6 \cr & \Rightarrow \frac{{{a^3} + {b^3} + {c^3} - 3abc}}{{16 - 10}} = 6 \cr & \Rightarrow {a^3} + {b^3} + {c^3} - 3abc = 6 \times 6 \cr & \Rightarrow {a^3} + {b^3} + {c^3} - 3abc = 36 \cr} $$