Đáp án: $2x^3+2x^2+1=2$
Giải thích các bước giải:
Ta có:
$x=\dfrac13\left(\sqrt[3]{\dfrac{23+\sqrt{513}}{4}}+\sqrt[3]{\dfrac{23-\sqrt{513}}{4}}-1\right)$
$\to 3x+1=\sqrt[3]{\dfrac{23+\sqrt{513}}{4}}+\sqrt[3]{\dfrac{23-\sqrt{513}}{4}}$
$\to \left(3x+1\right)^3=\left(\sqrt[3]{\dfrac{23+\sqrt{513}}{4}}+\sqrt[3]{\dfrac{23-\sqrt{513}}{4}}\right)^3$
$\to \left(3x+1\right)^3=\dfrac{23+\sqrt{513}}{4}+\dfrac{23-\sqrt{513}}{4}+3\cdot \sqrt[3]{\dfrac{23+\sqrt{513}}{4}}\cdot \sqrt[3]{\dfrac{23-\sqrt{513}}{4}}\left( \sqrt[3]{\dfrac{23+\sqrt{513}}{4}}+ \sqrt[3]{\dfrac{23-\sqrt{513}}{4}}\right)$
$\to \left(3x+1\right)^3=\dfrac{23}{2}+3\cdot \sqrt[3]{\dfrac{23^2-513}{16}}\left(3x+1\right)$
$\to \left(3x+1\right)^3=\dfrac{23}{2}+3\left(3x+1\right)$
$\to 27x^3+27x^2+9x+1=\dfrac{23}{2}+9x+3$
$\to 27x^3+27x^2=\dfrac{27}{2}$
$\to x^3+x^2=\dfrac{1}{2}$
$\to 2x^3+2x^2=1$
$\to 2x^3+2x^2+1=2$
