Question

(x+y)(x-y) + (y+z)(y-z) + (z+x)(z-x)=0​

Answer 1

Answer:

$\huge\underline\mathbb\red{AnSwEr}$

Step-by-step explanation:

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

x²-xy+xy-y²+y²-zy+zy-z²+z²-zx+zx-x²

=0

Answer 2

Given:

The term (x+y)(x-y)+(y+z)(y-z)+(z+x)(z-x)=0

To Find:

Prove the above term

Solution:

• Now we have given the term (x+y)(x-y)+(y+z)(y-z)+(z+x)(z-x)=0
• Let's consider LHS we have:
• (x+y)(x-y)+(y+z)(y-z)+(z+x)(z-x)
• Now we know the formula, which is:

(a-b)(a+b)=a^2-b^2

• So applying it in LHS, we get:

(x^2-y^2)+(y^2-z^2)+(z^2-x^2)

• Now adding it we get:

x^2-y^2+y^2-z^2+z^2-x^2

0.......RHS

Answer:

So in Solution part we proved that (x+y)(x-y)+(y+z)(y-z)+(z+x)(z-x)=0

$hope \: it \: will \: help \: u \\ have \:a \: nyc \: day$