Factorise: $(a^{2} + b^{2} - 4 c^{2})^{2} - 4 a^{2} b^{2}$

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

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

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

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Solution

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

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Similar questions

a, b, c

\frac{1}{2} {log}_{e} a + \frac{1}{2} {log}_{e} c + (\frac{1}{2 c a + 1}) + \frac{1}{3} {(\frac{1}{2 c a + 1})}^{3} + \frac{1}{5} {(\frac{1}{2 c a + 1})}^{5} + . . . \infty = {log}_{e} b

b

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

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

\frac{1}{a^{2} - b^{2}} + \frac{1}{b^{2}} = \frac{1}{b^{2} - c^{2}}

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

(a + b + 2 c)

If a^2+b^2+4c^2=ab+2bc+2ca,prove that a=b=2c.

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