?$ 解：(1)原式=x3+x2-\frac {1}{3}nx-3mx2-3mx+mn$?
?$=x3+(1-3m)x2-(\frac {1}{3}n+3m)x+mn$?
?$結(jié)果不含x2與x項(xiàng)，則1-3m=0，\frac {1}{3}n+3m=0$?
?$解得：m=\frac {1}{3}，n=-3$?
?$(2)原式=(3m-n)2+(\mathrm {mn})^{100}·n$?
?$=[3×\frac {1}{3}-(-3)]2+(-1)^{100}·(-3)$?
?$=16-3$?
?$=13$?
