Question

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

Answer 1

Answer:

${a}^{2} - {b}^{2} = (a + b)(a - b)$

$(x + y)(x - y) + \\ (y + z)(y - z) + \\ (z + x)(z - x) \\ \\ \\ = \cancel{x}^{2} - \cancel{y}^{2} + \cancel{y}^{2} - \cancel{z}^{2} + \cancel {z}^{2} \\ - \cancel{x}^{2} \\ \\ = 0$

Hence proved