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kap26 [50]
3 years ago
10

(x²-y²)³+(y²-z²)³+(z²-x²)³/(x-y)³+(y-z)³+(z-x)³ evaluate

Mathematics
1 answer:
svlad2 [7]3 years ago
6 0
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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