1.

1+2ab-(a²+b²) solve​

Answer»

Step-by-step explanation:

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)


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