Question

Show that: (x-y) (x+y) + (y-z) (y+z) + (z-x) (z+x) = 0​

Answer 1

Given that,

xy(x−y)+yz(y−z)+zx(z−x)

=(x×x×y)−(x×y×y)+(y×y×z)−(y×z×z)+(z×z×x)−(z×x×x)

→x

2

y−xy

2

+y

2

z−yz

2

+z

2

x−zx

2

Answer 2

Answer: 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

Lets 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