Find all roots of the equation $$2x^3+x^2+75x-81=0$$
Please check my work,
Let $P(x)$ : $2x^3+x^2+75x-81=0$
$ 2\left(x-\frac{1}{6}\right)^3 + \left(x-\frac{1}{6}\right)^2 + 75\left(x-\frac{1}{6}\right)-81 = 0$
$2x^3 + \frac{1}{6}x - \frac{1}{3}x +75x-\frac{1}{108}+\frac{1}{36}- \frac{75}{6}-81 = 0$
$2x^3 + 74\frac{5}{6}x - 93\frac{13}{27}=0$
$x^3 + 37\frac{5}{12}x - 46\frac{20}{27}=x^3+px+q = 0$
Let $u, v \in \mathbb{R}$ and $u+v$ be the real root of the equation.
By Cardano method,
$u^3v^3 = \frac{-\left(37\frac{5}{12}\right)^3}{27} = -\frac{p^3}{27}u^3+v^3 = 46\frac{20}{27} = -q$
$p=37\frac{5}{12}, q= -46\frac{20}{27}$
$\{u^3, v^3\}= \left\{\frac{-q}{2}-\frac{1}{2}\sqrt{q^2+\frac{4p^3}{27}}, \frac{-q}{2}+\frac{1}{2}\sqrt{q^2+\frac{4p^3}{27}}\right\}$
choose $u, v$ for which $uv = -\frac{p}{3}$
$\{u, v\}= \left\{\sqrt[3]{\frac{-q}{2}-\frac{1}{2}\sqrt{q^2+\frac{4p^3}{27}}}, \sqrt[3]{\frac{-q}{2}+\frac{1}{2}\sqrt[3]{q^2+\frac{4p^3}{27}}}\right\}$
Thus, roots of the equation are
$u+v-\frac{1}{6}, u\omega^2+v\omega-\frac{1}{6},u\omega+v\omega^2-\frac{1}{6} $ , where $\omega = cis\frac{2\pi}{3}$
$$2x^3+x^2+75x-81=2\left(x^3+\frac{1}{2}x^2+\frac{75}{2}x-\frac{81}{2}\right)=$$ $$=2\left(x^3+3\cdot\frac{1}{6}x^2+3\left(\frac{1}{6}\right)^2x+\left(\frac{1}{6}\right)^3+\left(\frac{75}{2}-\frac{1}{12}\right)x-\frac{1}{216}-\frac{81}{2}\right)=$$ $$=2\left(\left(x+\frac{1}{6}\right)^3+\frac{449}{12}x-\frac{8749}{216}\right)=$$ $$=2\left(\left(x+\frac{1}{6}\right)^3+\frac{449}{12}\left(x+\frac{1}{6}\right)-\frac{449}{72}-\frac{8749}{216}\right)=$$ $$=2\left(\left(x+\frac{1}{6}\right)^3+\frac{449}{12}\left(x+\frac{1}{6}\right)-\frac{1262}{27}\right).$$ Thus, we need to solve the following equation $$y^3+py+q=0,$$ where $y=x+\frac{1}{6}$, $p=\frac{449}{12}$ and $q=-\frac{1262}{27}$.
It's obvious that our equation has unique real root
because for example $$\left(\left(x+\frac{1}{6}\right)^3+\frac{449}{12}\left(x+\frac{1}{6}\right)\right)'=3\left(x+\frac{1}{6}\right)^2+\frac{449}{12}>0.$$ Let $y=u+v$ is a root of the equation $y^3+py+q=0$.
Hence, $$u^3+v^3+3uv(u+v)+p(u+v)+q=0$$ or $$u^3+v^3+q+(u+v)(3uv+p)=0.$$ Let $uv=-\frac{p}{3}$.
Hence, $u^3+v^3=-q$, which says that $u^3$ and $v^3$ are roots of the equation $$z^2+qz-\frac{p^3}{27}=0$$ or $$z^2-\frac{1262}{27}z-\frac{1}{27}\left(\frac{449}{12}\right)^3=0,$$ which gives $$x=-\frac{1}{6}+\sqrt[3]{\frac{631}{27}+\sqrt{\left(\frac{631}{27}\right)^2+\frac{1}{27}\left(\frac{449}{12}\right)^3}}+\sqrt[3]{\frac{631}{27}-\sqrt{\left(\frac{631}{27}\right)^2+\frac{1}{27}\left(\frac{449}{12}\right)^3}}$$