If x + y + z = 6 and x² + y² + z² = 20, then the value of x³ + y³ + z³ - 3xyz is?

Solution(By Examveda Team)

$$\eqalign{ & x + y + z = 6 \cr & {x^2} + {y^2} + {z^2} = 20 \cr & \Rightarrow {\left( {x + y + z} \right)^2} = {\left( 6 \right)^2} \cr & \Rightarrow {x^2} + {y^2} + {z^2} + 2\left( {xy + yz + zx} \right) = 36 \cr & \Rightarrow 20 + 2\left( {xy + yz + zx} \right) = 36 \cr & \Rightarrow 2\left( {xy + yz + zx} \right) = 16 \cr & \Rightarrow xy + yz + zx = 8 \cr & \therefore {\text{ }}{x^3} + {y^3} + {z^3} - 3xyz \cr & = \left( {x + y + z} \right)\left( {{\text{ }}{x^2} + {y^2} + {z^2} - xy - zx - yz} \right) \cr & = {x^3} + {y^3} + {z^3} - 3xyz = 6\left( {20 - 8} \right) \cr & = 6\left( {20 - 8} \right) \cr & = 6 \times 12 \cr & = 72 \cr} $$