7.(2023•葫芦岛二模)已知函数f(x)=ax3−ax−xlnxf(x)=ax^{3}-ax-xlnxf(x)=ax3−ax−xlnx,且f(x)⩾0f(x)\geqslant 0f(x)⩾0.(1)求aaa;(2)证明:f(x)f(x)f(x)存在唯一的极大值点x0x_{0}x0,且e−32<f(x0)<1e{e}^{-\frac{3}{2}}<f({x}_{0})<\frac{1}{e}e−23<f(x0)<e1.