Question

(x – y)(x + y) + (y – z)(y + z) + (z – x) (z + x) is equal to​

Answer 1

Answer:x (y-z)+y (z-x)+z (x-y)

=xy-xz+yz-yx+zx-zy

=(xy-yx)+(-xz+zx)+(yz-zy)

=0

Step-by-step explanation:If u like it pls mark me as brainlist

Answer 2

Given Data : variables x, y and z

To Find: $(x -y)(x + y) + (y - z)(y + z) + (z- x) (z + x)$ = ?

Solution:

• There are three variables x, y, z.

• Let us consider as A + B + C, A =$(x - y)(x + y)$,  $B = (y - z)(y + z)$  , $C= (z - x) (z + x)$ .

• Multiplying the values of $A = x^{2}+xy -xy - y^{2}$ ; $B=y^{2} +zy-zy-z^{2}$ ; $C=z^{2} +xz-xz -x^{2}$.

• Let us assume X.

• Substituting the values we get,
• X = A + B + C.
• X = $x^{2}+xy -xy - y^{2}$$+y^{2} +zy-zy-z^{2}$$z^{2} +xz-xz -x^{2}$.
• All the Values Cancel out each other.
• Hence $(x -y)(x + y) + (y - z)(y + z) + (z- x) (z + x) = 0$.