帮做做几个题，先给30分，都对的话+100分，急啊

1.a^4+(b+c)^4-2a^2(b+c)^2=[a^2-(b+c)^2]^2
=(a+b+c)^2*(a-b-c)^2
2.x^6-64y^6=x^6-(2y)^6
=(x^3+y^3)(x^3-y^3)
=(x+y)*(x^2-xy+y^2)*(x-y)(x^2+xy+y^2)
3.（x-y)(x-3y)+2y-1 =x^2+3*y^2-4xy+2y-1
=(x-2y)^2-(y-1)^2
=(x-y-1)*(x-3y+1)
4.a^2*(a+1)+b^2*(b+1)+ab(a-b)
=(a^3-b^3)+(a^2-b^2)+ab(a-b)
=(a-b)(a^2+ab+b^2)+(a+b)(a-b)+ab(a-b)
=(a-b)[(a+b)^2+(a+b)]
=(a-b)(a+b)(a+b+1)
5(a+b)^2+6(a^2-b^2)+9(a-b)^2
=a^2+b^2+2ab+6a^2-6b^2+9a^2+9b^2 -18ab
=16a^2-16ab+4b^2
=4(2a-b)^2
6.
设√(x^2-2x+10)=t(t≥0),则2x^2-4x=2*t^2-20
原方程变为
2*t^2-20-5t=5,解得t=5或t=-5/2(舍)
所以√(x^2-2x+10)=5，
解得
x1=-3,x2=5

(1)原式=a^4+(b+c)^4-2a^2(b+c)^@
=[a^2+(b+c)^2][a^2-(b+c)^2]
=[a^2+(b+c)^2](a+b+c)[a-(b+c)]
==[a^2+(b+c)^2](a+b+c)(a-b-c)

(2)原式=x^6-2^6*y^6
=(x^3+2^3*y^3)(x^3-2^3*y^3)
=(x+2y)(x^2-2xy+4y^