Factorize:
$a^{2} + b^{2} - c^{2} - d^{2} + 2 a b - 2 c d$ .

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

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

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

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

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Solution

Given:
$a^{2} + b^{2} - c^{2} - d^{2} + 2 a b - 2 c d$
$= a^{2} + b^{2} + 2 a b - c^{2} - d^{2} - 2 c d$
$= (a^{2} + b^{2} + 2 a b) - (c^{2} + d^{2} + 2 c d)$
$= (a + b)^{2} - (c + d)^{2}$
$= (a + b + c + d) (a + b - c - d)$ [ $∵ (x^{2} - y^{2} = (x + y) (x - y)$ ]
$∴ a^{2} + b^{2} - c^{2} - d^{2} + 2 a b - 2 c d = (a + b + c + d) (a + b - c - d)$
Hence, Option A is correct.

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Similar questions

∣ ∣ ∣ ∣ ∣ ∣ \begin{matrix} a & b & c & d b & a & d & c c & d & a & b d & c & b & a \end{matrix} ∣ ∣ ∣ ∣ ∣ ∣ = (a + b + c + d) (a - b + c - d) (a - b - c + d) (a + b - c - d)

If (a + b +c + d) (a -b- c+ d)

=(a + b- c- d) (a -b+c- d);

prove that : a : b =c : d.

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