Algebra 02

We will first start by finding the roots of the equation : $6x^2-6x+1=0$ and we do so by the "Quadratic formula" : $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$ where $a=6 , b=-6 , c=1$

we therefore get that the two roots are $\frac{3+\sqrt{3}}{6} , \frac{3-\sqrt{3}}{6}$ and we can say that $\alpha =\frac{3+\sqrt{3}}{6}, \beta =\frac{3-\sqrt{3}}{6}$ now the problem is simple , all we have to do is rearrange the equation , and equate coefficients : $\frac{1}{2}(a+b\alpha +c\alpha ^2+d\alpha ^3)+\frac{1}{2}(a+b\beta +c\beta ^2+d\beta ^3) =a+\frac{1}{2}b(\alpha +\beta )+\frac{1}{2}c(\alpha ^2+\beta^2 )+\frac{1}{2}d(\alpha ^3+\beta ^3)$ and this equation can be written as : $\frac{a}{p}+\frac{b}{q}+\frac{c}{r}+\frac{d}{s}$ therefore by equating coefficients one finds : $\frac{1}{p}=1 , \frac{1}{q}=\frac{\alpha +\beta }{2} , \frac{1}{r}=\frac{\alpha ^2+\beta ^2}{2},\frac{1}{s}=\frac{\alpha ^3+\beta ^3}{2}$ and by computing these values one finds : $p=1 , q=2 , r=3 , s =4$ Therefore $\frac{r.s}{p.q}=\frac{3*4}{1*2}=\frac{12}{2}=6$