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

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