(x²-y²)³+(y²-z²)³+(z²-x²)³/(x-y)³+(y-z)³+(z-x)³ evaluate
1 answer:
We use the fact that if x+y+z = 0, then x³+y³+ z³ = 3 x y z.<span>
(x</span>²-y²) + (y²-z²) + (z²-x²) = 0<span>
also: (x-y) + (y-z)+ (z-x) = 0
we assume that: x </span>≠y ≠ z.
<span>
hence,
(x²-y²)³ + (y²-z²)³ + (z²-x²)³ ÷ (x-y)³ + (y-z)³ + (z-x)³
= 3 (x</span>²-y²) (y²-z²) (z²-x²) ÷ [3 (x-y) (y-z) (z-x)]
<span>= (x+y) (y+z) (z+x)
</span>
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