A Challenge to all Level 5 Part 2

Let's make the substitution $x=\dfrac{y-1}{3}$ to delete the quadratic term:

$\left(\dfrac{y-1}{3}\right)^3+\left(\dfrac{y-1}{3}\right)^2+10\left(\dfrac{y-1}{3}\right)-3=0$

Expand and multiply by $27$ :

$y^3+87y-169=0$

Now, try to compare that equation with the identity $(u+v)^3-3uv(u+v)-(u^3+v^3)=0$ . An equation system will be formed:

$\quad \begin{aligned} y=u+v &\quad ...(1) \\ 3uv=-87 &\quad ...(2) \\ u^3+v^3=169 &\quad ...(3) \end{aligned}$ .

Divide $(2)$ by $3$ and cube both sides:

$u^3v^3=-24389$

Now, let's make an equation in $z$ with roots $u^3$ and $v^3$ :

$(z-u^3)(z-v^3)=0 \Longrightarrow z^2-(u^3+v^3)z+u^3v^3=0 \Longrightarrow z^2-169z-24389=0$

Using the quadratic formula, the solutions for $z$ are:

$z=\dfrac{169 \pm 27\sqrt{173}}{2}$

Hence, the real values of $u$ and $v$ are:

$u=\sqrt[3]{\dfrac{169+27\sqrt{173}}{2}}$ , $v=\sqrt[3]{\dfrac{169-27\sqrt{173}}{2}}$

There also two pairs of $(u,v)$ , they are $(uw,vw^2)$ and $(uw^2,vw)$ where $w$ is any primitive 3rd root of unity. But since we are asked for the real root, we only take these values.

Finally let's undo the substitutions:

$y=u+v=\sqrt[3]{\dfrac{169+27\sqrt{173}}{2}}+\sqrt[3]{\dfrac{169-27\sqrt{173}}{2}}$

$x=\dfrac{y-1}{3}=\dfrac{\sqrt[3]{\dfrac{169+27\sqrt{173}}{2}}+\sqrt[3]{\dfrac{169-27\sqrt{173}}{2}}-1}{3}$

$x \approx \boxed{0.2892}$

are u practicing cubics day and night