题意

$p=\sqrt[3] {4}-\sqrt[3] {6}+\sqrt[3] {9}$，求证：$1< p < 2$

证明:

$p\approx 1.8$，证毕(大雾

$1<p$挺显然的就不证明了

接下来证明$p<2$

$p<2$等价于$\sqrt[3] 9 -\sqrt[3] 8<\sqrt[3] 6-\sqrt[3] 4$

$9-8=(\sqrt[3]9-\sqrt[3]8)\times(\sqrt[3]9^2+\sqrt[3]8^2+\sqrt[3]{72})$

$6-4=(\sqrt[3]6-\sqrt[3]4)\times(\sqrt[3]6^2+\sqrt[3]4^2+\sqrt[3]{24})$

$\sqrt[3]9-\sqrt[3]8=\frac{1}{(\sqrt[3]9^2+\sqrt[3]8^2+\sqrt[3]{72})}=\frac{1}{(\sqrt[3]81+\sqrt[3]64+\sqrt[3]{72})}$

$\sqrt[3]6-\sqrt[3]4=\frac{1}{(\sqrt[3]6^2+\sqrt[3]4^2+\sqrt[3]{24})}=\frac{1}{(\sqrt[3]36+\sqrt[3]16+\sqrt[3]{24})}$

$(\sqrt[3]81+\sqrt[3]64+\sqrt[3]{72})>(\sqrt[3]36+\sqrt[3]16+\sqrt[3]{24})$

$\frac{1}{(\sqrt[3]81+\sqrt[3]64+\sqrt[3]{72})}<\frac{1}{(\sqrt[3]36+\sqrt[3]16+\sqrt[3]{24})}$