Question : Simplify the given expression. $\frac{x^3+y^3+z^3-3 x y z}{(x-y)^2+(y-z)^2+(z-x)^2}$
Option 1: $\frac{1}{3}(x+y+z)$
Option 2: $(x+y+z)$
Option 3: $\frac{1}{4}(x+y+z)$
Option 4: $\frac{1}{2}(x+y+z)$
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Correct Answer: $\frac{1}{2}(x+y+z)$
Solution :
By polynomial identity
$x^3+y^3+z^3−3xyz = (x+y+z)(x^2+y^2+z^2−xy−yz−zx)$
= $\frac{1}{2}(x+y+z)(2x^2+2y^2+2z^2−2xy−2yz−2zx)$
= $\frac{1}{2}(x+y+z)(x^2−2xy+y^2+y^2−2yz+z^2+z^2−2xz+x^2)$
= $\frac{1}{2}[(x+y+z)(x−y)^2+(y−z)^2+(z−x)^2]$
$\therefore \frac{x^3+y^3+z^3-3 x y z}{(x-y)^2+(y-z)^2+(z-x)^2}$ = $\frac{\frac{1}{2}[(x+y+z)(x−y)^2+(y−z)^2+(z−x)^2]}{(x-y)^2+(y-z)^2+(z-x)^2}$ = $\frac{1}{2}(x+y+z)$
Hence, the correct answer is $\frac{1}{2}(x+y+z)$.
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