Question

Factorise 1 - 2ab - (a 2 + b 2 ) .

Answer 1

Factorization of the expression $1-2 a b-\left(a^{2}+b^{2}\right)$ is  

$\bold{(1+a+b) (1-a-b)}$

Given:

$1-2 a b-\left(a^{2}+b^{2}\right)$          

To Find:

Factorization of $1-2 a b-\left(a^{2}+b^{2}\right)$

Solution:

The given expression is,  

$1-2 a b-\left(a^{2}+b^{2}\right)$

Now, the above expression can be written as  

$1-\left(a^{2}+b^{2}+2 a b\right)$

The above expression can be written as,

$=1-(a+b)^{2}\left[\because(a+b)^{2}=a^{2}+b^{2}+2 a b\right]$

Now, we get

$=(1)^{2}-(a+b)^{2}\left[\text { Since, } 1=1^{2}\right]$

Now, on expanding the above equation, the new expression is,

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

Now, the equation becomes,

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

Answer 2

Answer:

Factors of 1-2ab-(a²+b²)

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

Explanation:

Given 1-2ab-(a²+b²)

= 1-2ab-a²-b²

= 1-(2ab+a²+b²)

= 1-(a+b)²

/* By algebraic identity : x²+y²+2xy =(x+y)² */

= 1²-(a+b)²

= [1+(a+b)][1-(a+b)]

/* By algebraic identity :

x²-y² =(x+y)(x-y) */

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

Therefore,

1-2ab-(a²+b²) = (1+a+b)(1-a-b)

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