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Sử dụng bất đẳng thức $(a+b)^3 =a^{3} +b^3 +3ab(a+b)$Ta có.$x=\sqrt[3]{\frac{27+\sqrt{713}}{108}}+\sqrt[3]{\frac{27-\sqrt{713}}{108}} - \frac{1}{3}$$\Leftrightarrow x+\frac{1}{3}= \sqrt[3]{\frac{27+\sqrt{713}}{108}}+\sqrt[3]{\frac{27-\sqrt{713}}{108}}$$\Leftrightarrow (x+\frac{1}{3})^3=(\sqrt[3]{\frac{27+\sqrt{713}}{108}}+\sqrt[3]{\frac{27-\sqrt{713}}{108}})^3$$\Leftrightarrow (x+\frac{1}{3})^3= \frac{27+\sqrt{713}}{108}+\frac{27-\sqrt{713}}{108}+3\sqrt[3]{\frac{27+\sqrt{713}}{108}}.\sqrt[3]{\frac{27-\sqrt{713}}{108}}(x+\frac{1}{3} )$ $\Leftrightarrow (x+\frac{1}{3})^3=\frac{1}{2}+ 3.\frac{1}{9}(x+\frac{1}{3} )$$\Leftrightarrow x^3+x^2+\frac{1}{3}x+\frac{1}{27} =\frac{1}{2}+ \frac{1}{3}x+\frac{1}{9}$$\Leftrightarrow x^3+x^2-1 = \frac{-23}{54}$

Sử dụng hằng đẳng thức $(a+b)^3 =a^{3} +b^3 +3ab(a+b)$Ta có.$x=\sqrt[3]{\frac{27+\sqrt{713}}{108}}+\sqrt[3]{\frac{27-\sqrt{713}}{108}} - \frac{1}{3}$$\Leftrightarrow x+\frac{1}{3}= \sqrt[3]{\frac{27+\sqrt{713}}{108}}+\sqrt[3]{\frac{27-\sqrt{713}}{108}}$$\Leftrightarrow (x+\frac{1}{3})^3=(\sqrt[3]{\frac{27+\sqrt{713}}{108}}+\sqrt[3]{\frac{27-\sqrt{713}}{108}})^3$$\Leftrightarrow (x+\frac{1}{3})^3= \frac{27+\sqrt{713}}{108}+\frac{27-\sqrt{713}}{108}+3\sqrt[3]{\frac{27+\sqrt{713}}{108}}.\sqrt[3]{\frac{27-\sqrt{713}}{108}}(x+\frac{1}{3} )$ $\Leftrightarrow (x+\frac{1}{3})^3=\frac{1}{2}+ 3.\frac{1}{9}(x+\frac{1}{3} )$$\Leftrightarrow x^3+x^2+\frac{1}{3}x+\frac{1}{27} =\frac{1}{2}+ \frac{1}{3}x+\frac{1}{9}$$\Leftrightarrow x^3+x^2-1 = \frac{-23}{54}$