The factorization form of $a^{4} + b^{4} - a^{2} b^{2}$ is _____________.

Open in App

Solution

We have $a^{4} + b^{4} - a^{2} b^{2}$
Adding and subtracting $2 a^{2} b^{2},$ we get
$= (a^{2})^{2} + (b^{2})^{2} + 2 a^{2} b^{2} - 2 a^{2} b^{2} - a^{2} b^{2}$
$= (a^{2} + b^{2})^{2} - 3 a^{2} b^{2}$ [Using the identity: $x^{2} + y^{2} + 2 x y = (x + y)^{2}$ ]
$= (a^{2} + b^{2})^{2} - 3 a^{2} b^{2}$
$= (a^{2} + b^{2} + \sqrt{3} a b) (a^{2} + b^{2} - \sqrt{3} a b)$ [Using the identity: $(a^{2} - b^{2}) = (a + b) (a - b)$ ]
Hence, the factorization form of $a^{4} + b^{4} - a^{2} b^{2}$ is $(a^{2} + b^{2} + \sqrt{3} a b) (a^{2} + b^{2} - \sqrt{3} a b)$

Suggest Corrections

Similar questions

a^{4} + b^{4} + c^{4} \geq a^{2} b^{2} + b^{2} c^{2} + c^{2} a^{2}

a^{4} + b^{4} = a^{2} b^{2}

(a^{6} + b^{6})

Δ A B C

c^{4} - 2 (a^{2} + b^{2}) c^{2} + a^{4} + a^{2} b^{2} + b^{4} = 0

∠ C =

a^{2} + a b + b^{2} = 19, a^{4} + a^{2} b^{2} + b^{4} = 133,

b

a

c

\frac{a^{4} + a^{2} b^{2} + b^{4}}{b^{4} + b^{2} c^{2} + c^{4}} = \frac{a^{2}}{c^{2}}

Join BYJU'S Learning Program