If x = 999, y = 1000, z = 1001 then the value of $\frac{x^{3} + y^{3} + z^{3} - 3 x y z}{x - y + z}$ $\frac{{{x^3} + {y^3} + {z^3} - 3xyz}}{{x - y + z}}$ is?

11000
29000
31
49

Answer:- 1

Explanation:-

Solution:

\begin{aligned} ∴ a^{3} + b^{3} + c^{3} - 3 a b c \\ = \frac{1}{2} (a + b + c) [{(a - b)}^{2} + {(b - c)}^{2} + {(c - a)}^{2}] \\ ∴ \frac{x^{3} + y^{3} + z^{3} - 3 x y z}{x - y + z} \\ = \frac{\frac{1}{2} (x + y + z) [{(x - y)}^{2} + {(y - z)}^{2} + {(z - x)}^{2}]}{x - y + z} \\ = \frac{\frac{1}{2} (999 + 1000 + 1001) (1 + 1 + 4)}{999 - 1000 + 1001} \\ = \frac{\frac{1}{2} \times 6 \times 3000}{1000} \\ = 9 \end{aligned}

$\eqalign{ & \therefore {a^3} + {b^3} + {c^3} - 3abc \cr & = \frac{1}{2}\left( {a + b + c} \right)\left[ {{{\left( {a - b} \right)}^2} + {{\left( {b - c} \right)}^2} + {{\left( {c - a} \right)}^2}} \right] \cr & \therefore \frac{{{x^3} + {y^3} + {z^3} - 3xyz}}{{x - y + z}} \cr & = \frac{{\frac{1}{2}\left( {x + y + z} \right)\left[ {{{\left( {x - y} \right)}^2} + {{\left( {y - z} \right)}^2} + {{\left( {z - x} \right)}^2}} \right]}}{{x - y + z}} \cr & = \frac{{\frac{1}{2}\left( {999 + 1000 + 1001} \right)\left( {1 + 1 + 4} \right)}}{{999 - 1000 + 1001}} \cr & = \frac{{\frac{1}{2} \times 6 \times 3000}}{{1000}} \cr & = 9 \cr}$