Question

x square minus y square whole cube + y square minus Z Square whole cube + Z square minus y square whole cube by x minus y whole cube + y - x whole cube + Z - X whole cube​

Answer 1

hope this helps you.. :)

Answer 2

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

Explanation:

•From the given expression consider Numerator:

(x²-y²)³+(y²-z²)³+(z²-x²)³

•Consider (x²-y²) = A,(y²-z²) = B,(z²-x²) = C

•Let A+B+C = x²-y²+y²-z²+z²-x² = 0

•If A+B+C = 0,then (A³+B³+C³) = 3ABC

• therefore,numerator = 3(x²-y²)(y²-z²)(z²-x²)

• similarly,denominator (x-y)³+(y-z)³+(z-x)³ = 3(x-y)(y-z)(z-x)

•Therefore,Numerator/Denominator= (3(x²-y²)(y²-z²)(z²-x²))/(3(x-y)(y-z)(z-x))

= ((x²-y²)(y²-z²)(z²-x²))/((x-y)(y-z)(z-x)) ..(1)

• we know that (a²-b²) = (a+b)(a-b)

• from the above formula (x²-y²) can be written as (x+y)(x-y)

• similarly, equation(1) can be written as

((x+y)(y+z)(z+x)(x-y)(y-z))(z-x)/((x-y)(y-z)(z-x))

• Therefore,

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