Question 142

If $$x = 2, y = 1$$ and $$z = -3$$, then $$x^3 + y^3 + z^3 - 3xyz$$ is equal to

Solution

Given, $$x$$ = 2, $$y$$ = 1, $$z$$ = -3 then

$$x$$ + $$y$$ + $$z$$ = 2 + 1 - 3 = 0

When $$x$$ + $$y$$ + $$z$$ = 0 then $$x^3+y^3+z^3=3xyz$$

Therefore $$x^3+y^3+z^3-3xyz=0$$

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