题目你能化简(x−1)(x99+x98+…+…+x+1)\left(x-1\right)(x^{99}+x^{98}+\ldots +\ldots +x+1)(x−1)(x99+x98+…+…+x+1)吗?遇到这样的复杂问题时,我们可以先从简单的情形入手.然后归纳出一些方法.(1)(1)(1)分别化简下列各式:(x−1)(x+1)=(x-1)\left(x+1\right)=(x−1)(x+1)=______;(x−1)(x2+x+1)=(x-1)(x^{2}+x+1)=(x−1)(x2+x+1)=______;(x−1)(x3+x2+x+1)=(x-1)(x^{3}+x^{2}+x+1)=(x−1)(x3+x2+x+1)=______;…\ldots…(x−1)(x99+x98+…+x+1)=______.(x-1)(x^{99}+x^{98}+\ldots +x+1)=\_\_\_\_\_\_.(x−1)(x99+x98+…+x+1)=______.(2)(2)(2)请你利用上面的结论计算:299+298+…+2+12^{99}+2^{98}+\ldots +2+1299+298+…+2+1.知识点:多项式乘多项式章节:数与式 / 整式