(x^2减1/x)^6的二项展开式中含x^6的系数是过程及方法 急

(x^2减1/x)^6 =(x²-1/x)³(x²-1/x)³ =(x^6-3x³+3-1/x³)(x^6-3x³+3-1/x³) =x^12-3x^9+3x^6-x³-3x^9+9x^6-9x^3+3+3x^6-9x^3+9-3/x^3-x^3+3-3/x^3+1/x^6 =x^12-6x^9+15x^6-20x^3+18-6/x^3+