Factorize: $(a^{2} - 1) (b^{2} - 1) + 4 a b$

(a b + 1 + a + b) (a b + 1 - a + b)

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

(a b + 1 + a + b) (2 a b + 1 - a + b)

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

(2 a b + 1 + a + b) (a b + 1 + a + b)

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

(2 a b + 1 + a + b) (2 a b + 1 + a + b)

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

Open in App

Solution

The correct option is A $(a b + 1 + a + b) (a b + 1 - a + b)$
$(a^{2} - 1) (b^{2} - 1) + 4 a b$
$= a^{2} b^{2} - a^{2} - b^{2} + 1 + 2 a b + 2 a b$
$= (a b)^{2} + 2 (a b) (1) + 1^{2} - (a^{2} - 2 a b + b^{2})$
$= (a b + 1)^{2} - (a - b)^{2}$
$= (a b + 1 + a - b) (a b + 1 - a + b)$

Suggest Corrections

Similar questions

l o g_{6}^{30} = a, l o g_{15}^{24} = b t h e n l o g_{12}^{60} =

A = [\begin{matrix} 2 & - 1 0 & 1 \end{matrix}]

B = [\begin{matrix} 1 & 0 - 1 & - 1 \end{matrix}],

(A + B)^{2} = A^{2} + A B + B A + B^{2} = A^{2} + 2 A B + B^{2}

∣ ∣ ∣ ∣ ∣ \begin{matrix} 1 + a^{2} - b^{2} & 2 a b & - 2 b 2 a b & 1 - a^{2} + b^{2} & 2 a 2 b & - 2 a & 1 - a^{2} - b^{2} \end{matrix} ∣ ∣ ∣ ∣ ∣ =

A = (\begin{matrix} 1 & - 1 2 & 1 \end{matrix}), B = (\begin{matrix} a & 1 b & - 1 \end{matrix})

(A + B)^{2} = A^{2} + B^{2} + 2 A B

Join BYJU'S Learning Program