原式=x^12(x³+1)+x^6(x³+1)+（x³+1）
=(x³+1)(x^12+x^6+1)
=(x+1)(x²-x+1)(x^12+x^6+1)
=(x+1)(x²-x+1)(x^12+2x^6+1-x^6)
=(x+1)(x²-x+1)[(x^6+1)²-x^6]
=(x+1)(x²-x+1)(x^6+x³+1)(x^6-x³+1)
請點擊下面的【選為滿意回答】按鈕,