已知3x的3次方－x=1,求9x的4次方＋12x的3次方－3x的平方－7x＋1999的值

3x³-x=1
原式=(9x^4-3x²)+12x³-7x+2001
=3x(3x³-x)+12x³-7x+2001
=3x×1+12x³-7x+2001
=3x+12x³-7x+2001
=12x³-4x+2001
=4(3x³-x)+2001
=4×1+2001
=2005