Giải cho mình bài 2 với ạ

Đáp án+Giải thích các bước giải:

 `a)(x+y)^{2}-(x-y)^{2}=(x+y-x+y).(x+y+x-y)=2y.2x=4xy`

`b)(a+b)^{3}+(a-b)^{3}-2a^{3}`

`=a^{3}+3a^{2}b+3ab^{2}+b^{3}+a^{3}-3a^{2}b+3ab^{2}-b^{3}-2a^{3}`

`=(a^{3}+a^{3}-2a^{3})+(3a^{2}b-3a^{2}b)+(3ab^{2}+3ab^{2})+(b^{3}-b^{3})`

`=6ab^{2}`

`c)(x+3).(x+7)-(x+1).(x-1)`

`=x.(x+7)+3.(x+7)-(x^{2}-1^{2})`

`=x^{2}+7x+3x+21-x^{2}+1`

`=(x^{2}-x^{2})+(7x+3x)+(21+1)`

`=10x+22`

`d)(x+1).(x^{2}-x+1)-(x-1).(x^{2}+x+1)`

`=(x+1).(x^{2}-x.1+1^{2})-(x-1).(x^{2}+x.1+1^{2})`

`=x^{3}+1^{3}-(x^{3}-1^{3})`

`=x^{3}-x^{3}+1+1`

`=2`

`e)(2+1).(2^{2}+1).(2^{4}+1).(2^{8}+1).(2^{16}+1)`

`=(2^{2}-1).(2^{2}+1).(2^{4}+1).(2^{8}+1).(2^{16}+1)`

`=(2^{4}-1).(2^{4}+1).(2^{8}+1).(2^{16}+1)`

`=(2^{8}-1).(2^{8}+1).(2^{16}+1)`

`=(2^{16}-1).(2^{16}+1)`

`=(2^{16})^{2}-1^{2}`

`=2^{32}-1`