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>
Answer:
.
Step-by-step explanation:
(x+2)(x+4)
=x(x+4)+2(x+4)
=x²+4x+2x+8
=x²+6x+8
Answer:
The answer is 18
Step-by-step explanation:
Combine like terms by adding their FACES (Coefficients)
1) 4c² - 4
2) 20m^4 + 6m
3) -6a³ - 19a² + 8a - 10
4) -10a² - 2b + 3ab³ + 4a²b² -5b^4 - 9a²b² - 12a³b + 15ab³ = -10a² - 2b + 18ab³ - 5a²b² - 12a³b - 5b^4
5) 5a³ - 5a²
