factorize 2x+4y+x^3+8y^3

2x + 4y + x^3 + 8y^3

= 2x + 4y + ( x + 2y )( x^2 - 2xy + 4y^2)

= 2( x + 2y ) + ( x + 2y )( x^2 - 2xy + 4y^2)

= ( x + 2y ) [ 2 + ( x^2 - 2xy + 4y^2) ]

= ( x + 2y ) ( x^2 - 2xy + 4y^2 + 2 )

2007-10-04 11:45:16 補充：
利用恆等式 a^3 + b^3 ≡ ( a + b )( a^2– ab + b^2 )

since a3 + b3 = (a+b) (a2 -2ab +b2)
therefore 2x+4y+x3+8y3
= 2x+4y+x3+(2y)3
= 2x+4y+(x+2y)(x2-2xy+4y2)
=(x+2y)(x2-2xy+4y2+2)

2x+4y+x^3+8y^3
= 2(x+2y) + (x+2y)(x^2-2xy+4y^2)
= (x+2y)(x^2-2xy+4y^2+2)