If x + y + z = 6 and xy + yz + zx = 10, then the value of x³ + y³ + z³ - 3xyz is?

Answer:- 1

Explanation:-

Solution:

\begin{aligned} Given \\ x + y + z = 6 \\ x y + y z + z x = 10 \\ To find x^{3} + y^{3} + z^{3} - 3 x y z =  ? \\ \Rightarrow Using formula, \\ \Rightarrow {(x + y + z)}^{2} = x^{2} + y^{2} + z^{2} + 2 (x y + y z + z x) \\ \Rightarrow 6^{2} = x^{2} + y^{2} + z^{2} + 2 \times 10 \\ \Rightarrow 36 = x^{2} + y^{2} + z^{2} + 20 \\ \Rightarrow x^{2} + y^{2} + z^{2} = 16 \\ \Rightarrow x^{2} + y^{2} + z^{2} - 3 x y z = (x + y + z) (x^{2} + y^{2} + z^{2} - x y - y z - z x) \\ \Rightarrow x^{2} + y^{2} + z^{2} - 3 x y z = 6 [16 - 10] \\ \Rightarrow x^{2} + y^{2} + z^{2} - 3 x y z = 6 \times 6 \\ \Rightarrow x^{2} + y^{2} + z^{2} - 3 x y z = 36 \end{aligned}

$\eqalign{ & {\text{Given}} \cr & x + y + z = 6 \cr & xy + yz + zx = 10 \cr & {\text{To find }}{x^3} + {y^3} + {z^3} - 3xyz{\text{ = ?}} \cr & \Rightarrow {\text{ Using formula,}} \cr & \Rightarrow {\left( {x + y + z} \right)^2} = {x^2} + {y^2} + {z^2} + 2\left( {xy + yz + zx} \right) \cr & \Rightarrow {6^2} = {x^2} + {y^2} + {z^2} + 2 \times 10 \cr & \Rightarrow 36 = {x^2} + {y^2} + {z^2} + 20 \cr & \Rightarrow {x^2} + {y^2} + {z^2} = 16 \cr & \Rightarrow {x^2} + {y^2} + {z^2} - 3xyz = \left( {x + y + z} \right)\left( {{x^2} + {y^2} + {z^2} - xy - yz - zx} \right) \cr & \Rightarrow {x^2} + {y^2} + {z^2} - 3xyz = 6\left[ {16 - 10} \right] \cr & \Rightarrow {x^2} + {y^2} + {z^2} - 3xyz = 6 \times 6 \cr & \Rightarrow {x^2} + {y^2} + {z^2} - 3xyz = 36 \cr}$