Briot (an English mathematician, who lived from 1817 to 1882) and Ruffini (an Italian mathematician, who lived from 1765 to 1822) developed methods to find solutions for equations called reciprocal. In this question, you will develop, step by step, the essence of these methods. Item (a) is a preparation for the other items.

(a) If $y=x+\frac{1}{x}$, calculate the expressions $x^{2}+\frac{1}{x^{2}}$ and $x^{3}+\frac{1}{x^{3}}$ in terms of $y$.

(b) Determine all real roots of the equation $x^{2}-5 x+8-\frac{5}{x}+\frac{1}{x^{2}}=0$.

(c) Determine all real roots of $x^{4}-5 x^{3}+8 x^{2}-5 x+1=0$.

(d) Determine all real roots of $x^{6}-2 x^{5}-5 x^{4}+12 x^{3}-5 x^{2}-2 x+1=0$.
(a) We have $y=x+\frac{1}{x}$. Using the binomial and trinomial expansions, we get

$$
\begin{aligned}
& y^{2}=\left(x+\frac{1}{x}\right)^{2}=x^{2}+2 x \frac{1}{x}+\frac{1}{x^{2}}=x^{2}+\frac{1}{x^{2}}+2 \\
& y^{3}=\left(x+\frac{1}{x}\right)^{3}=x^{3}+3 x^{2} \frac{1}{x}+3 x \frac{1}{x^{2}}+\frac{1}{x^{3}}=x^{3}+3\left(x+\frac{1}{x}\right)+\frac{1}{x^{3}}
\end{aligned}
$$

therefore,

$$
x^{2}+\frac{1}{x^{2}}=y^{2}-2 \quad \text { and } \quad x^{3}+\frac{1}{x^{3}}=y^{3}-3\left(x+\frac{1}{x}\right)=y^{3}-3 y
$$

(b) The given equation is equivalent to $\left(x^{2}+\frac{1}{x^{2}}\right)-5\left(x+\frac{1}{x}\right)+8=0$. Substituting the value of $y$ and using the identity from the previous item, we get

$$
\left(y^{2}-2\right)-5 y+8=0
$$

that is, the quadratic equation $y^{2}-5 y+6=0$, whose roots are $y=2$ and $y=3$. Returning to $x$, we multiply $x+\frac{1}{x}=y=2$ and $y=3$ by $x$ to obtain the quadratic equations $(x-1)^{2}=x^{2}-2 x+1=0$ and $x^{2}-3 x+1=0$, whose roots are $x=1$ and $x=\frac{1}{2}(3 \pm \sqrt{5})$. Thus, we have obtained all three roots of the given equation.

(c) Since $x=0$ is not a root of the given equation, we can divide everything by $x^{2}$. In this way, we find exactly the equation from item (b), whose roots we have already obtained.

(d) Since $x=0$ is not a root of the given equation, we can divide everything by $x^{3}$. In this way, rearranging the terms, we obtain the equivalent equation

$$
\left(x^{3}+\frac{1}{x^{3}}\right)-2\left(x^{2}+\frac{1}{x^{2}}\right)-5\left(x+\frac{1}{x}\right)+12=0
$$

Substituting the value of $y$ and using the identities from item (a), we get

$$
\left(y^{3}-3 y\right)-2\left(y^{2}-2\right)-5 y+12=0
$$

equivalent to the cubic equation

$$
y^{3}-2 y^{2}-8 y+16=0
$$

The fastest way to solve this equation is to have a bit of luck and factor by grouping, for example,

$$
y^{3}-2 y^{2}-8 y+16=y^{2}(y-2)-8(y-2)=\left(y^{2}-8\right)(y-2)
$$

so the three roots of the cubic equation in $y$ are $y=2$ and $y= \pm 2 \sqrt{2}$. Returning to $x$, we multiply $x+\frac{1}{x}=y=2$ and $y= \pm 2 \sqrt{2}$ by $x$ to obtain the quadratic equations $(x-1)^{2}=x^{2}-2 x+1=0, x^{2}-2 \sqrt{2} x+1=0 \text{ and } x^{2}+2 \sqrt{2} x+1=0$, whose roots are $x=1, x=\sqrt{2} \pm 1$ and $x=-\sqrt{2} \pm 1$. Thus, we have obtained all five roots of the given equation.