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Question Number 131363 by shaker last updated on 04/Feb/21

Commented by MJS_new last updated on 04/Feb/21

$$\mathrm{I}{\mathrm{don}}^{\prime }\mathrm{t}\mathrm{think}\mathrm{we}\mathrm{can}\mathrm{exactly}\mathrm{solve}\mathrm{this}.\mathrm{we}$$$$\mathrm{need}\mathrm{the}\mathrm{roots}\mathrm{of}{x}^3+3x+1\mathrm{and}\mathrm{in}\mathrm{the}\mathrm{next}$$$$\mathrm{step}\mathrm{the}\mathrm{roots}\mathrm{of}\mathrm{a}\mathrm{polynome}\mathrm{of}\mathrm{degree}4\mathrm{if}$$$$\mathrm{we}\mathrm{are}\mathrm{smart}...\mathrm{otherwise}\mathrm{degree}6$$

Commented by MJS_new last updated on 04/Feb/21

$$\mathrm{try}\mathrm{these}\mathrm{to}\mathrm{see}\mathrm{the}\mathrm{path}$$$$\left(1\right)x^3+3x+1=(x-a)(x-b)(x-c)$$$$[\mathrm{in}\mathrm{our}\mathrm{case}a\in \mathbb{R};b,c\notin \mathbb{R}]$$$$\left(2\right)x^3+3x+1=(x-a)(x^2+ax-\frac{1}{a})$$$$[a\in \mathbb{R}]$$$$\mathrm{in}\mathrm{both}\mathrm{cases}\mathrm{you}\mathrm{can}\mathrm{use}$$$$t=x+\sqrt{x^2+1}\iff x=\frac{t^2-1}{2t}\to dx=\frac{\sqrt{x^2+1}}{t}dt$$$${\mathrm{it}}^{\prime }\mathrm{s}\mathrm{possible}\mathrm{to}\mathrm{solve}\mathrm{it}\mathrm{but}\mathrm{exactly}?\mathrm{no}.$$

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