Question

14. Prove that
(a) (x + y)(x - y) + (y + z)(y – z) + (z + x)(z - x) = 0​

Answer 1

Step-by-step explanation:

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

x²-y²+y²-z²+z²-x²= 0

therefore LHS=RHS

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Answer 2

Step-by-step explanation:

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

Formula a²-b²=(a+b)(a-b)

x²-y²+y²-z²+z²-x²=0

0=0

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