Question

The value of (x-y)(x+y)+(y-z)(y+z)+(z-x)(z+x) is :

a) x+y+z
b) x-y+yz+zx
c) 0​

Answer 1

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

Answer 2

Step-by-step explanation:

x = a.(b-c). ………….(1).

y = b.(c-a). ……………..(2).

z = c.(a-b). ………………(3).

Adding the eqn.(1) , (2) and (3).

x + y + z = ab -ca+bc -ab +ca -bc.

or, x + y + z = 0. ……………..(4).

xy(x+y)+yz.(y+z)+zx.(z+x) +3xyz.=?

Putting (x+y)=-z , (y+z) = -x and (z+x) = -y from eqn.(4).

= xy(-z) +yz(-x)+zx(-y) +3xyz.

= -3xyz + 3xyz.

= 0. Answer.