Find the value of $(a - b) (a + b) + (b - c) (b + c) + (c - a) (c + a)$ .

a b c

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a^{2} + b^{2} + c^{2}

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0

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1

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Solution

The correct option is B $0$
We know, $(a - b) (a + b) = a^{2} - b^{2}$ .

$∴ (a - b) (a + b) + (b - c) (b + c) + (c - a) (c + a)$
$= (a^{2} - b^{2}) + (b^{2} - c^{2}) + (c^{2} - a^{2})$
$= a^{2} - b^{2} + b^{2} - c^{2} + c^{2} - a^{2}$
$= 0$ .

Therefore, option $C$ is correct.

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Similar questions

\frac{a^{2} - b^{2} - c^{2}}{(a - b) (a - c)} + \frac{b^{2} - c^{2} - a^{2}}{(b - c) (b - a)} + \frac{c^{2} - a^{2} - b^{2}}{(c - a) (c - b)}

a, b, c

A, B, C \in R - (0}

a A + b B + c C + \sqrt{(a^{2} + b^{2} + c^{2}) (A^{2} + B^{2} + C^{2})} = 0

\frac{a B}{b A} + \frac{b C}{c B} + \frac{c A}{a C}

△_{1}, △_{2}

(b, c), (c, a), (a, b)

(a c - b^{2}, a b - c^{2}), (b a - c^{2}, b c - a^{2}), (c b - a^{2}, c a - b^{2})

\frac{△_{1}}{△_{2}} = (a + b + c)^{2}

\frac{a^{2}}{b c} + \frac{b^{2}}{a c} + \frac{c^{2}}{b a}

a + b + c = 0

\frac{(b + c)^{2}}{b c} + \frac{(c + a)^{2}}{c a} + \frac{(a + b)^{2}}{a b}

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