Factorize:
x3+y3+z3=3xyz.
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3
Answer:
We have x
3
+y
3
+z
3
−3xyz
=(x+y+z)(x
2
+y
2
+z
2
−xy−yz−zx)
0=(x+y+z)(x
2
+y
2
+z
2
−xy−yz−zx)
So either x+y+z=0 (or)
x
2
+y
2
+z
2
−xy−yz−zx=0
x
2
+y
2
+z
2
−xy−yz−zx=0
2x
2
+2y
2
+2z
2
−2xy−2yz−2zx=0
(x
2
−2xy+y
2
)+(y
2
−2yz+z
2
)+(z
2
−2zx+x
2
)=0
(x−y)
2
+(y−z)
2
+(z−x)
2
=0
⇒x=y=z
Therefore, either x+y+z=0 or x=y=z.
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