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

1 answer:

6 0

We use the fact that if x+y+z = 0, then x³+y³+ z³ = 3 x y z.

(x²-y²) + (y²-z²) + (z²-x²) = 0
also: (x-y) + (y-z)+ (z-x) = 0
we assume that: x ≠y ≠ z.


hence,

(x²-y²)³ + (y²-z²)³ + (z²-x²)³ ÷ (x-y)³ + (y-z)³ + (z-x)³
= 3 (x²-y²) (y²-z²) (z²-x²) ÷ [3 (x-y) (y-z) (z-x)]
= (x+y) (y+z) (z+x)

You might be interested in

( 20 points + brainliest for the correct answer!!!) ASA AAS CONGRUENCE

Olegator [25]

A. AAS
B. SAS
C. ASA

Just look at what order they come in. Lmk if you need further help.

5 0

2 years ago

Find the GCF. 14abc and 28 a 2 b 2 c 3

9966 [12]

4 because 14abc would equal 108 the factors of 108 are 1,2,3,4,6,9,12,18,27,36,54,108 the factors of 28 are 1,2,4,7,14,28 so it's 4

7 0

3 years ago

Which of the following functions are solutions of the differential equation y'' + y = 3 sin x? (select all that apply)

DiKsa [7]

Answer:

Step-by-step explanation:

We must compute the derivatives and check if the equation is satisfied.

A. $y'=(3\sin x)'=3\cos x$ . Differentiate again to get $y''=(3\cos x)'=-3\sin x$ , then $y''+y=-3\sin x+3\sin x=0\neq 3\sin x$ so this choice of y doesn't solve the equation.

B. $y'=3\sin x+3x\cos x-5\cosx +5x\sin x=(5x+3)\sin x+(3x-5)\cos x$ and $y''=5\sin x+(5x+3)\cos x+3\cos x-(3x-5)\sin x=(10-3x)\sin x+(5x+6)\cos x$ , then $y''+y=10\sin x+6\cos x\neq 3\sin x$ so y is not a solution

C. $y'=\frac{-3}{2}\cos x+\frac{3}{2}x\sin x$ hence $y''=\frac{3}{2}\sin x+\frac{3}{2}\sin x+\frac{3}{2}x\cos x=3\sin x+\frac{3}{2}x\cos x$ . Then $y''+y=3\sin x+\frac{3}{2}x\cos x-\frac{3}{2}x\cos x=3\sin x$ so y is a solution.

D. $y'=-3\sin x$ and $y''=-3\cos x$ , then $y''+y=0$ thus y isn't a solution

E. $y'=\frac{3}{2}\sin x+\frac{3}{2}x\cos x$ hence $y''=\frac{3}{2}\cos x+\frac{3}{2}\cos x-\frac{3}{2}x\sin x=3\cos x-\frac{3}{2}x\cos x$ . Then $y''+y=3\cos x-\frac{3}{2}x\cos x-\frac{3}{2}x\cos x=3\cos x\neq 3\sin x$ then y is not a solution.