Factorize the equation ${(a^{2} + b^{2} - 4 c^{2})}^{2} - 4 a^{2} b^{2}$

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

No worries! We‘ve got your back. Try BYJU‘S free classes today!

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

No worries! We‘ve got your back. Try BYJU‘S free classes today!

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

No worries! We‘ve got your back. Try BYJU‘S free classes today!

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

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

Solution

The correct option is D $(a - b + 2 c) (a - b - 2 c) (a + b + 2 c) (a + b - 2 c)$
${(a^{2} + b^{2} - 4 c^{2})}^{2} - 4 a^{2} b^{2}$
$= {(a^{2} + b^{2} - 4 c^{2})}^{2} - (2 a b)^{2}$
$= [a^{2} + b^{2} - 4 c^{2} + 2 a b] [a^{2} + b^{2} - 4 c^{2} - 2 a b]$
Using,
$(a + b)^{2} = a^{2} + 2 a b + b^{2}$ and $(a - b)^{2} = a^{2} - 2 a b + b^{2}$
$= [(a + b)^{2} - 4 c^{2}] [(a - b)^{2} - 4 c^{2}]$
Using,
$a^{2} - b^{2} = (a + b) (a - b)$
$= (a + b + 2 c) (a + b - 2 c) (a - b - 2 c) (a - b + 2 c)$

Suggest Corrections

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

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

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.

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

Join BYJU'S Learning Program