Question

factorise x - y whole cube + y - Z whole cube + Z - X whole cube

Answer 1

Answer:

$(x-y)^3+(y-z)^3+(z-x)^3=0$  

Step-by-step explanation:

Given expression $(x-y)^3+(y-z)^3+(z-x)^3$  ........(1)

We have to factorize it.

We know, $(a-b)^3= a^3-3a^2b +3ab^2-b^3$

Applying to given expression,

$(x-y)^3=x^3-3x^2y+3xy^2-y^3\\\\(y-z)^3=y^3-3y^2z+3yz^2-z^3\\\\(z-x)^3=z^3-3z^2x+3x^2z-x^3$

Put in (1) , we get,

$x^3-3x^2y+3xy^2-y^3+y^3-3y^2z+3yz^2-z^3+z^3-3z^2x+3x^2z-x^3$

Cancelling similar terms, we get,

$(x-y)^3+(y-z)^3+(z-x)^3=0$  

Answer 2

"To determine: The factors of the given expression

Given: Expression: ${(x-y)}^{3} +{(y-z)}^{3} +{(z-x)}^{}$

Formula used: ${ (a-b) }^{ 3 } = {a}^{3}- 3{a}^{2}b+ 3a{b}^{2}- {b}^{3}$

Calculation:

We know that if (a+b+c) = 0 then ${a}^{3} + {b}^{3} + {c}^{3} = 3abc$

So, if we assign a = x-y, b = y-z and c = z-x

a+ b+ c = x-y + y-z + z-x = 0

Thus ${(x-y)}^{3} +{(y-z)}^{3} + {(z-x)}^{3} = 3(x-y) (y-z) (z-x)$"