Question

Factorise : 1 - 2ab - (a²+ b²)
​

Answer 1

Answer:

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

Given:

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

1−2ab−(a 2 +b 2 )

To Find:

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

2

+b

2

)

Solution:

The given expression is,

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

2

+b

2

)

Now, the above expression can be written as

1-\left(a^{2}+b^{2}+2 a b\right)1−(a

2

+b

2

+2ab)

The above expression can be written as,

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

2

[∵(a+b)

2

=a

2

+b

2

+2ab]

Now, we get

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

2

−(a+b)

2

[ Since, 1=1

2

]

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

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

Now, the equation becomes,

=(1+a+b)(1-a-b)=(1+a+b)(1−a−b)

Answer 2

Answer:

Identity (a²+b²) = a² + b² + 2ab

Step-by-step explanation:

• 1 - 2ab - (a² + b²)
• = 1 - 2ab - [(a)² + (b)² + 2ab ]
• = 1 - 2ab - a² - b² - 2ab
• = 1 - a² - b² - 4ab

Hope it helps you..