Question

Show that

(i) (a – b) (a + b) + (b – c) (b + c) + (c – a) (c + a) = 0

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Answer 1

Step-by-step explanation:

$(a - b)(a + b) + (b - c)(b + c) + (c - a)(c + a) \\ \\ = a {}^{2} - b {}^{2} + b {}^{2} - c {}^{2} + c {}^{2} - a {}^{2} \\ \\ = 0 \: \: \: \: (proved) \\ \\ \\ formula \: \: \: \: - \: \: \: \: (x + y)(x - y) = x {}^{2} - y {}^{2}$

Answer 2

Answer:

a square - b square + b square - C square + c square - a square = 0. ( (a+b)(a-b) = square - b square)

(arrange it in like terms)

a square - a square + b square - b square + c square - C square= 0

0+0+0=0

0=0

LHS =RHS

HP