## Solve the equation for $$x\epsilon\mathbb{R}$$ (Cubic equation Cardano’s method)

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To solve the equation, let’s start by multiplying both sides by $$\left(x-1\right)\left(x-2\right)$$ and simplifying, we get $\frac{x}{x-1}\left(x-1\right)\left(x-2\right)=\left(x+\frac{1}{x-2}\right)\left(x-1\right)\left(x-2\right)\;\;\;\;\;\;\;,\;x\neq1\;and\;x\neq2$ $\Rightarrow x\left(x-2\right)=x\left(x-1\right)\left(x-2\right)+x-1$ $\Rightarrow x^{2}-2x=x^{3}-2x^{2}-x^{2}+2x+x-1$ $\Rightarrow x^{3}-4x^{2}+5x-1=0$ Let $$f\left(x\right)=x^{3}-4x^{2}+5x-1$$.  Now we are dealing with a cubic equation, this can be made by replacing $$x$$ by $$y+\frac{4}{3}$$ to get rid of $$x^{2}$$ term, this gives $\left(y+\frac{4}{3}\right)^{3}-4\left(y+\frac{4}{3}\right)^{2}+5\left(y+\frac{4}{3}\right)-1=0$ $y^{3}+3y^{2}\frac{4}{3}+3y\left(\frac{4}{3}\right)^{2}+\left(\frac{4}{3}\right)^{3}-4\left(y^{2}+2y\frac{4}{3}+\left(\frac{4}{3}\right)^{2}\right)+5y+\frac{20}{3}-1=0$ $y^{3}+4y^{2}+\frac{16}{3}y+\frac{64}{27}-4y^{2}-\frac{32}{3}y-\frac{64}{9}+5y+\frac{20}{3}-1=0$ $\Rightarrow y^{3}-\frac{1}{3}y+\frac{25}{27}=0$ We get a depressed cubic equation, to solve it we assume that $$y=s-t$$ is a solution to the equation $$y^{3}+Ay+B=0$$ with $$A=-\frac{1}{3}$$ and $$B=\frac{25}{27}$$, we get $\left(s-t\right)^{3}+A\left(s-t\right)+B=0$ $s^{3}-3s^{2}t+3st^{2}-t^{3}+As-At+B=0$ $B+s^{3}-t^{3}-s\left(3st-A\right)+t\left(3st-A\right)=0$ $\Rightarrow B+s^{3}-t^{3}+\left(A-3st\right)\left(s-t\right)=0$
At this point we impose the condition $$A-3st=0$$ following Cardano’s method, this helps to remove the fourth term in the previous equality, therefore we obtain a system of equations, we get $B+s^{3}-t^{3}=0$ $A-3st=0$ Writting $$s=\frac{A}{3t}$$ and using the first equation of the previous system, we get $B+\left(\frac{A}{3t}\right)^{3}-t^{3}=0$ To solve that, let’s make a variable change using $$u=t^{3}$$, we get $B+\frac{A^{3}}{27u}-u=0$ Multiplying by $$u$$ with $$u\neq0$$ gives a quadratic equation $u^{2}-Bu-\frac{A^{3}}{27}=0$ Let’s substitute $$A$$ and $$B$$ with their values, we get $u^{2}-\frac{25}{27}u+\frac{1}{729}=0$ Solving for $$u$$ gives $u=\frac{25+3\sqrt{69}}{54}$ Here we used only the larger solution, the other solution also works. This gives $t=\frac{1}{3}\sqrt{\frac{25+3\sqrt{69}}{2}}$ $and$ $s=\frac{A}{3t}=-\frac{1}{3}\sqrt{\frac{2}{25+3\sqrt{69}}}$ Hence, the solution $x=y+\frac{4}{3}=s-t+\frac{4}{3}$ $x=\frac{4}{3}-\frac{1}{3}\sqrt{\frac{2}{25+3\sqrt{69}}}-\frac{1}{3}\sqrt{\frac{25+3\sqrt{69}}{2}}\approx0.24$ Now, let’s be sure that it is the only solution. To make this, we calculate the derivative $f’\left(x\right)=3x^{2}-8x+5=\left(x-1\right)\left(3x-5\right)$ It is positive in $$\left(-\infty,1\right)$$, negative in $$\left(1,\frac{5}{3}\right)$$ and positive in $$\left(\frac{5}{3},\infty\right)$$. Since $$f\left(1\right)=1$$ and $$f\left(\frac{5}{3}\right)=\frac{23}{27}$$, the function $$f$$ is positive in $$\left[1,\infty\right)$$ and has only one zero which is $$x\approx0.24$$ in $$\left(-\infty,1\right)$$.
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