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Question Number 55501 by ajfour last updated on 25/Feb/19

$$Rootsof{x}^3+px+q=0$$$$arex=u+v$$$$u^3,v^{3}arerootsof$$$$z^2-{\alpha }^{3}z+{\beta }^6=0$$$$\frac{d^2\left(y/x\right)}{{dx}^2}{\mid }_{x=\alpha }=0,\frac{dy}{dx}{\mid }_{x=\beta }=0.$$$$Ihavenoticedlongwayback,$$$$pleasehuntwhy\left(howcome\right)?$$

Commented by ajfour last updated on 25/Feb/19

$$\frac{dy}{dx}=3x^2+p=0\Rightarrow {\beta }^6=-\frac{p^3}{27}$$$$\frac{y}{x}=x^2+p+\frac{q}{x}$$$$\frac{d\left(y/x\right)}{dx}=2x-\frac{q}{x^2}$$$$\frac{d^2\left(y/x\right)}{{dx}^2}=2+\frac{2q}{x^3}=0\Rightarrow {\alpha }^3=-q$$$$z^2-{\alpha }^{3}z+{\beta }^6=0$$$$\Rightarrow z_1,z_2=u^3,v^{3}are$$$$=\frac{{\alpha }^3}{2}\pm \sqrt{{\left(\frac{{\alpha }^3}{2}\right)}^2-{\beta }^6}$$$$=-\frac{q}{2}\pm \sqrt{\frac{q^2}{4}+\frac{p^3}{27}}$$$$x=u+v$$$$={(-\frac{q}{2}+\sqrt{\frac{q^2}{4}+\frac{p^3}{27}})}^{1/3}$$$$+{(-\frac{q}{2}+\sqrt{\frac{q^2}{4}+\frac{p^3}{27}})}^{1/3}.$$$$Justhavenoticedso!$$

Commented by mr W last updated on 25/Feb/19

$$welcomebacksir!$$

Commented by ajfour last updated on 25/Feb/19

$$wasbusywithpolynomialsSir,$$$$foundnothingnew,didlotof$$$$experimenting.$$

Commented by mr W last updated on 25/Feb/19

$$gladtoknowthatthingsaregoing$$$$wellwithyousir!$$

Commented by ajfour last updated on 25/Feb/19

$$thanksfortheconcernSir.$$

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