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

Roots of x^3 +px+q=0 are x = u+v u^3 , v^3 are roots of z^2 −α^3 z+β^6 =0 ((d^2 (y/x))/dx^2 )∣_(x=α) =0 , (dy/dx)∣_(x=β) =0 . I have noticed long way back, please hunt why (how come) ?

$${Roots}\:{of}\:{x}^{\mathrm{3}} +{px}+{q}=\mathrm{0} \\ $$$${are}\:\:{x}\:=\:{u}+{v} \\ $$$${u}^{\mathrm{3}} ,\:{v}^{\mathrm{3}} \:{are}\:{roots}\:{of} \\ $$$$\:\:\:\:\:\:\:{z}^{\mathrm{2}} −\alpha^{\mathrm{3}} {z}+\beta^{\mathrm{6}} =\mathrm{0} \\ $$$$\:\:\:\frac{{d}^{\mathrm{2}} \left({y}/{x}\right)}{{dx}^{\mathrm{2}} }\mid_{{x}=\alpha} =\mathrm{0}\:\:,\:\frac{{dy}}{{dx}}\mid_{{x}=\beta} =\mathrm{0}\:. \\ $$$${I}\:{have}\:{noticed}\:{long}\:{way}\:{back}, \\ $$$${please}\:{hunt}\:{why}\:\left({how}\:{come}\right)\:? \\ $$

Commented by ajfour last updated on 25/Feb/19

(dy/dx)=3x^2 +p =0 ⇒ β^( 6) = −(p^3 /(27)) (y/x)= x^2 +p+(q/x) ((d(y/x))/dx)=2x−(q/x^2 ) ((d^2 (y/x))/dx^2 )=2+((2q)/x^3 ) =0 ⇒ α^3 = −q z^2 −α^3 z+β^( 6) =0 ⇒ z_1 , z_2 = u^3 , v^3 are = (α^3 /2)±(√(((α^3 /2))^2 −β^( 6) )) = −(q/2)±(√((q^2 /4)+(p^3 /(27)))) x = u+v = (−(q/2)+(√((q^2 /4)+(p^3 /(27)))) )^(1/3) +(−(q/2)+(√((q^2 /4)+(p^3 /(27)))) )^(1/3) . Just have noticed so!

$$\frac{{dy}}{{dx}}=\mathrm{3}{x}^{\mathrm{2}} +{p}\:=\mathrm{0}\:\:\Rightarrow\:\:\beta^{\:\mathrm{6}} \:=\:−\frac{{p}^{\mathrm{3}} }{\mathrm{27}}\: \\ $$$$\frac{{y}}{{x}}=\:{x}^{\mathrm{2}} +{p}+\frac{{q}}{{x}} \\ $$$$\frac{{d}\left({y}/{x}\right)}{{dx}}=\mathrm{2}{x}−\frac{{q}}{{x}^{\mathrm{2}} } \\ $$$$\frac{{d}^{\mathrm{2}} \left({y}/{x}\right)}{{dx}^{\mathrm{2}} }=\mathrm{2}+\frac{\mathrm{2}{q}}{{x}^{\mathrm{3}} }\:=\mathrm{0}\:\:\Rightarrow\:\alpha^{\mathrm{3}} \:=\:−{q} \\ $$$$\:\:{z}^{\mathrm{2}} −\alpha^{\mathrm{3}} {z}+\beta^{\:\mathrm{6}} =\mathrm{0} \\ $$$$\Rightarrow\:\:{z}_{\mathrm{1}} ,\:{z}_{\mathrm{2}} \:=\:{u}^{\mathrm{3}} ,\:{v}^{\mathrm{3}} \:\:{are} \\ $$$$\:\:\:\:\:\:\:=\:\frac{\alpha^{\mathrm{3}} }{\mathrm{2}}\pm\sqrt{\left(\frac{\alpha^{\mathrm{3}} }{\mathrm{2}}\right)^{\mathrm{2}} −\beta^{\:\mathrm{6}} } \\ $$$$\:\:\:\:\:\:=\:−\frac{{q}}{\mathrm{2}}\pm\sqrt{\frac{{q}^{\mathrm{2}} }{\mathrm{4}}+\frac{{p}^{\mathrm{3}} }{\mathrm{27}}} \\ $$$$\:{x}\:=\:{u}+{v} \\ $$$$\:\:\:\:\:=\:\left(−\frac{{q}}{\mathrm{2}}+\sqrt{\frac{{q}^{\mathrm{2}} }{\mathrm{4}}+\frac{{p}^{\mathrm{3}} }{\mathrm{27}}}\:\right)^{\mathrm{1}/\mathrm{3}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+\left(−\frac{{q}}{\mathrm{2}}+\sqrt{\frac{{q}^{\mathrm{2}} }{\mathrm{4}}+\frac{{p}^{\mathrm{3}} }{\mathrm{27}}}\:\right)^{\mathrm{1}/\mathrm{3}} \:. \\ $$$${Just}\:{have}\:{noticed}\:{so}! \\ $$

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

welcome back sir!

$${welcome}\:{back}\:{sir}! \\ $$

Commented by ajfour last updated on 25/Feb/19

was busy with polynomials Sir, found nothing new, did lot of experimenting.

$${was}\:{busy}\:{with}\:{polynomials}\:{Sir}, \\ $$$${found}\:{nothing}\:{new},\:{did}\:{lot}\:{of} \\ $$$${experimenting}. \\ $$

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

glad to know that things are going well with you sir!

$${glad}\:{to}\:{know}\:{that}\:{things}\:{are}\:{going} \\ $$$${well}\:{with}\:{you}\:{sir}! \\ $$

Commented by ajfour last updated on 25/Feb/19

thanks for the concern Sir.

$${thanks}\:{for}\:{the}\:{concern}\:{Sir}. \\ $$

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