Question Number 131363 by shaker last updated on 04/Feb/21
![](https://www.tinkutara.com/question/18497.png)
Commented by MJS_new last updated on 04/Feb/21
![I don′t think we can exactly solve this. we need the roots of x^3 +3x+1 and in the next step the roots of a polynome of degree 4 if we are smart... otherwise degree 6](https://www.tinkutara.com/question/Q131367.png)
$$\mathrm{I}\:\mathrm{don}'\mathrm{t}\:\mathrm{think}\:\mathrm{we}\:\mathrm{can}\:\mathrm{exactly}\:\mathrm{solve}\:\mathrm{this}.\:\mathrm{we} \\ $$$$\mathrm{need}\:\mathrm{the}\:\mathrm{roots}\:\mathrm{of}\:{x}^{\mathrm{3}} +\mathrm{3}{x}+\mathrm{1}\:\mathrm{and}\:\mathrm{in}\:\mathrm{the}\:\mathrm{next} \\ $$$$\mathrm{step}\:\mathrm{the}\:\mathrm{roots}\:\mathrm{of}\:\mathrm{a}\:\mathrm{polynome}\:\mathrm{of}\:\mathrm{degree}\:\mathrm{4}\:\mathrm{if} \\ $$$$\mathrm{we}\:\mathrm{are}\:\mathrm{smart}…\:\mathrm{otherwise}\:\mathrm{degree}\:\mathrm{6} \\ $$
Commented by MJS_new last updated on 04/Feb/21
![try these to see the path (1) x^3 +3x+1=(x−a)(x−b)(x−c) [in our case a∈R; b, c ∉R] (2) x^3 +3x+1=(x−a)(x^2 +ax−(1/a)) [a∈R] in both cases you can use t=x+(√(x^2 +1)) ⇔ x=((t^2 −1)/(2t)) → dx=((√(x^2 +1))/t)dt it′s possible to solve it but exactly? no.](https://www.tinkutara.com/question/Q131368.png)
$$\mathrm{try}\:\mathrm{these}\:\mathrm{to}\:\mathrm{see}\:\mathrm{the}\:\mathrm{path} \\ $$$$\left(\mathrm{1}\right)\:{x}^{\mathrm{3}} +\mathrm{3}{x}+\mathrm{1}=\left({x}−{a}\right)\left({x}−{b}\right)\left({x}−{c}\right) \\ $$$$\:\:\:\:\:\left[\mathrm{in}\:\mathrm{our}\:\mathrm{case}\:{a}\in\mathbb{R};\:{b},\:{c}\:\notin\mathbb{R}\right] \\ $$$$\left(\mathrm{2}\right)\:{x}^{\mathrm{3}} +\mathrm{3}{x}+\mathrm{1}=\left({x}−{a}\right)\left({x}^{\mathrm{2}} +{ax}−\frac{\mathrm{1}}{{a}}\right) \\ $$$$\:\:\:\:\:\left[{a}\in\mathbb{R}\right] \\ $$$$\mathrm{in}\:\mathrm{both}\:\mathrm{cases}\:\mathrm{you}\:\mathrm{can}\:\mathrm{use} \\ $$$${t}={x}+\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}\:\Leftrightarrow\:{x}=\frac{{t}^{\mathrm{2}} −\mathrm{1}}{\mathrm{2}{t}}\:\rightarrow\:{dx}=\frac{\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}}{{t}}{dt} \\ $$$$\mathrm{it}'\mathrm{s}\:\mathrm{possible}\:\mathrm{to}\:\mathrm{solve}\:\mathrm{it}\:\mathrm{but}\:\mathrm{exactly}?\:\mathrm{no}. \\ $$