（2x-1）的七次方是多少?

Question

（2x-1）的七次方是多少?

Answer 1

=[(2x-1)^2]^2[(2x-1)^2(2x-1)]
=(4x^2-4x+1)^2[(4x^2-4x+1)(2x-1)]
=(16x^4+16x^2+1-32x^3+8x^2-8x)(8x^3-4x^2-8x^2+4x+2x-1)
=(16x^4-32x^3+24x^2-8x+1)(8x^3-12x^2+6x-1)
=128x^7-192x^6+96x^5-16x^4-256x^6+384x^5-192x^4+32x^3+192x^5-288x^4+144x^3-24x^2-64x^4+96x^3-48x^2+8x+8x^3-12x^2+6x-1
=128x^7-448x^6+672x^5-560x^4+280x^3-84x^2+14x-1