$a^{2} - b^{2} = (a + b) (a - b)$ . Then factors of $1 + 2 a b - (a^{2} + b^{2})$ are

Open in App

Solution

given, $1 + 2 a b - (a^{2} + b^{2})$

$= 1 + 2 a b - a^{2} - b^{2}$

$= 1 - (a^{2} + b^{2} - 2 a b)$

$= 1 - (a - b)^{2}$

$= (1)^{2} - (a - b)^{2}$

$= (1 + a - b) (1 - a + b)$

Suggest Corrections

Similar questions

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

1 + 2 a b - (a^{2} + b^{2})

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}

Simplify $\frac{a - b}{a} + \frac{a + b}{b}$

Factor(s) of $a^{2} - 2 a b + b^{2} - c^{2}$ is/are

Join BYJU'S Learning Program