Show that
$(a - b) (a + b) + (b - c) (b + c) + (c - a) (c + a) = 0$

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Solution

We know that,
$(a - b) (a + b) = a^{2} - b^{2}$

$\Rightarrow$ $(a - b) (a + b) + (b - c) (b + c) + (c - a) (c + a)$

$= a^{2} - b^{2} + b^{2} - c^{2} + c^{2} - a^{2}$

$= 0$

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

Question 5(v)
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(a-b)(a+b)+(b-c)(b+c)+(c-a)(c+a)=0

a (a - b) (a - c) + b (b - c) (b - a) + c (c - a) (c - b)

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

(i) {(3 x + 7)}^{2} - 84 x = {(3 x - 7)}^{2}

(i i) {(4 p q + 3 q)}^{2} - {(4 p q - 3 q)}^{2} = 48 p q^{2}

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

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