Factorise: $a^{4} + 4 b^{4} + 3 a^{2} b^{2}$

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

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

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

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

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

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

(a^{2} + 3 a b + 2 b^{2}) (a^{2} - 6 a b + 2 b^{2})

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

Open in App

Solution

The correct option is C $(a^{2} + a b + 2 b^{2}) (a^{2} - a b + 2 b^{2})$
$a^{4} + 4 b^{4} + 3 a^{2} b^{2}$
$= (a^{2})^{2} + (2 b^{2})^{2} + 2 a^{2} (2 b^{2}) - (a b)^{2}$
$= (a^{2} + 2 b^{2})^{2} - (a b)^{2}$
$= (a^{2} + 2 b^{2} + a b) (a^{2} + 2 b^{2} - a b)$

Suggest Corrections

Similar questions

\frac{7 a^{2} + 2 b^{2}}{7 a^{2} + 2 b^{2}} = \frac{113}{13}

\frac{a}{b}

a, b

c

\frac{a}{c} =

State True or False:

On subtracting $a^{2} + a b + b^{2}$ from $4 a^{2} - 3 a b + 2 b^{2}$ , the answer is $3 a^{2} - 4 a b + b^{2}$ .

a \neq b

2 a^{2} = 7 a - 5

2 b^{2} = 7 b - 5

a^{2} + b^{2}

If the roots of the equation $(a^{2} - b c) x^{2} + 2 (b^{2} - a c) x + c^{2} - a b = 0$ are equal, then

Join BYJU'S Learning Program