(2y+1)(2y-1)(4y的平方+1)怎么算?
问题描述：
(2y+1)(2y-1)(4y的平方+1)怎么算?
答
题目要干什么？化简后是16^4-1
答
(2y+1)(2y-1)(4y^2+1)=(4y^2-1)(4y^2+1)=16y^4-1;
(2y+1)(2y-1)=4y^2-1是平方差公式（a+b)(a-b)=a^2-b^2
答
(2y+1)(2y-1)(4y的平方+1)
=(4y²-1)(4y²+1)
=16y的4次方-1