Finding root of $x^3-36x^2+405x-1458=0$

Question

How can I solve the following equation :

$$x^3-36x^2+405x-1458=0$$

I have tried as :

$$x^3-36x^2+405x-1458=0$$ $$\Rightarrow x^2(x-36)+405(x-3.6)=0$$

How is to proceed ?

Answer 1

There are lots of powers of $3$ in there. Try $x=3y$, then $$27y^3-324y^2+1215y-1458=0\\y^3-12y^2+45y-54=0$$ Can you find a solution to that, or change it again?

Answer 2

Hint: Starting by looking for integer solutions. Every integer solution will necessarily be a divisor of $1458=2\cdot 3^6$.

Answer 3

Because $x=9$ is a solution, you can factor out $(x-9)$. Can you continue?

Answer 4

If we replace $x$ with $y+12$, then $y$ with $3z$, the original problem boils down to solving: $$ z^3-3z-2=0,$$ whose solutions are obviously given by $z=-1$ and $z=2$, since $z^3-3z+2=(z+1)^2(z-2)$.