Question

show that
(a-b)(a+b)+(b-c)(b+c)+(c-a)(c+a)​

Answer 1

Answer:

Hence, proved................

Answer 2

Answer:

(a-b)(a+b)+(b-c)(b+c)+(c-a)(c+a) = 0

Now we have to simplify equation within both side in equal range.

→ (LHS = RHS )

(a-b)(a+b)+(b-c)(b+c)+(c-a)(c+a) = 0

using the formula based on that equation

(a-b) (a+b) = a² - b² ,

(a-b)(a+b)+(b-c)(b+c)+(c-a)(c+a) = 0

→ a²- b²+ b²- c²+ c² - a² = 0

→ a²- a²- b² + b²- c² + c² = 0

→ 0 = 0

→RHS = LHS

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