The factors of $a^{2} - b^{2} - 4 c^{2} + 4 d^{2} - 4 (a d - b c)$ are:

(a + 2 d + b + 2 c) (a - 2 d - b + 2 c)

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(a - 2 d + b - 2 c) (a + 2 d - b + 2 c)

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(a - 2 d + b - 2 c) (a - 2 d - b + 2 c)

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(a - 2 d - b - 2 c) (a + 2 d + b + 2 c)

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Solution

The correct option is C $(a - 2 d + b - 2 c) (a - 2 d - b + 2 c)$
$a^{2} - b^{2} - 4 c^{2} + 4 d^{2} - 4 (a d - b c)$
$= a^{2} - b^{2} - 4 c^{2} + 4 d^{2} - 4 a d + 4 b c$
$= (a^{2} - 4 a d + 4 d^{2}) - (b^{2} - 4 b c + 4 c^{2})$
$= {(a - 2 d)}^{2} - {(b - 2 c)}^{2}$
$= (a - 2 d + b - 2 c) (a - 2 d - b + 2 c)$

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Similar questions

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

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

a^{2} + b^{2} + 4 c^{2} + 2 (a b + 2 b c + 2 c a)

(a + b + 2 c)

If a, b, c are in G.P., prove that:

(i) $a (b^{2} + c^{2}) = c (a^{2} + b^{2})$

(ii) $a^{2} b^{2} c^{2} (\frac{1}{a^{3}} + \frac{1}{b^{3}} + \frac{1}{c^{3}}) = a^{3} + b^{3} + c^{3}$

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

(iv) $\frac{1}{a^{2} - b^{2}} + \frac{1}{b^{2}} = \frac{1}{b^{2} - c^{2}}$

(v) $(a + 2 b + 2 c) (a - 2 b + 2 c) = a^{2} + 4 c^{2}$ .

a^{8} - 2 a^{4} b^{4} + b^{8}

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