Question 65

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

Solution

Given : x=999 y=1000, z=1001

Let $$y=k$$, => $$x=(k-1)$$ and $$z=(k+1)$$

To find : $$\frac{x^{3}+y^{3}+z^3-3xyz}{x-y+z}$$

= $$\frac{(k-1)^3+(k)^3+(k+1)^3-3[(k-1)(k)(k+1)]}{(k-1)-(k)+(k+1)}$$

= $$\frac{1}{k} \times [(k^3-1-3k^2+3k)+(k^3)+(k^3+1+3k^2+3k)-3(k^3-k)]$$

= $$\frac{1}{k} \times [(3k^3+6k)+(-3k^3+3k)]$$

=> $$\frac{9k}{k}=9$$

=> Ans - (D)

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