x的平方+x-1=0 求证(x+1)的立方-(x-1)的立方=8-6x
问题描述：
x的平方+x-1=0 求证(x+1)的立方-(x-1)的立方=8-6x
答
∵(X+1)³-(X-1)³=(X+1-X+1)[(X+1)²+(X+1)(X-1)+(X-1)²]
=2(X²+2X+1+X²-1+X²-2X+1)
=2(3X²+1)
=6X²+2
∵X²+X-1=0
∴X²=1-X
∴6X²+2=6(1-X)+2
=6-6X+2
=8-6X
答
(x+1)³-(x-1)³ =[(x+1)-(x-1)][(x+1)²+(x+1)(x-1)+(x-1)²] =2(3x²+1) 由已知 x²+x-1=0得 x²=1-x 代入上式得 原式=2(3x²+1) =2(3-3x+1) =8-6x...