Math, asked by ishanisharma36, 1 month ago

Factorise the following: x4-y4+x2-y2​

Answers

Answered by 070307
3

Answer:

Let us solve x

4

+y

4

+x

2

y

2

=(x

4

+y

4

+2x

2

y

2

)−x

2

y

2

=(x

2

+y

2

)

2

−(xy)

2

Using a

2

−b

2

=(a+b)(a−b)

x

4

+y

4

+x

2

y

2

=(x

2

+y

2

+xy)(x

2

+y

2

−xy)

Hence, option C is correct

Answered by shaiksubhanuddin123
0

Answer:

x

4

−y

4

+x

2

−y

2

</p><p>=(x+y)(x−y)[(x

2

+y

2

+1]

Step-by-step explanation:

\begin{gathered}x^{4}-y^{4}+x^{2}-y^{2}\\=[(x^{2})^{2}-(y^{2})^{2})]+x^{2}-y^{2}\\=(x^{2}+y^{2})(x^{2}-y^{2})+x^{2}-y^{2}\end{gathered}

x

4

−y

4

+x

2

−y

2

=[(x

2

)

2

−(y

2

)

2

)]+x

2

−y

2

=(x

2

+y

2

)(x

2

−y

2

)+x

2

−y

2

\* By algebraic identity :

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

a

2

−b

2

=(a+b)(a−b)

*\

=(x^{2}-y^{2})[(x^{2}+y^{2}+1](x

2

−y

2

)[(x

2

+y

2

+1]

=(x+y)(x-y)[(x^{2}+y^{2}+1](x+y)(x−y)[(x

2

+y

2

+1]

Therefore,

\begin{gathered}x^{4}-y^{4}+x^{2}-y^{2}\\ < /p > < p > =(x+y)(x-y)[(x^{2}+y^{2}+1]\end{gathered}

x

4

−y

4

+x

2

−y

2

</p><p>=(x+y)(x−y)[(x

2

+y

2

+1]

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