Question Number 128007 by ajfour last updated on 03/Jan/21
![Can anyone find any error in my attempt to improve upon the Cardano′s cubic formula.. or as an alternate to the trigonometric solution... here it goes: X^3 −pX−q=0 let (X/( (√p))) = x ; and with (q/(p(√p))) = c , x^3 −x−c=0 let x=(k+1)(p+q) =(p+kq) + (kp+q) p^3 +k^3 q^3 +3kpq(p+kq)+ k^3 p^3 +q^3 +3kpq(kp+q) −(p+kq)−(kp+q)=c ⇒ (1+k^3 )(p^3 +q^3 )+ (p+kq)(3kpq−1)+ (kp+q)(3kpq−1)= c let 3kpq=1 ⇒ p^3 +q^3 =(c/(1+k^3 )) & p^3 q^3 =(1/(27k^3 )) ; hence p^3 , q^3 = (c/(2(1+k^3 )))±(√((c^2 /(4(1+k^3 )^2 ))−(1/(27k^3 )))) lets choose upon a value of k such that D=0 ⇒ 4(1+k^3 )^2 =27c^2 k^3 .....(I) (just a quadratic..) first for c^2 >(8/(27)) we always can get two real k values, and even p=q then. x=(k+1)(p+q) = 2(k+1)p x=2(k+1)[(c/(2(1+k^3 )))]^(1/3) but simply pq=(1/(3k)) hence x=2(k+1)p = ((2(k+1))/( (√(3k)))) . ________________________ even for c=1 i dint get a correct answer, please help error-freeing it. (Thanks!)](Q128007.png)
$${Can}\:{anyone}\:{find}\:{any}\:{error}\:{in} \\ $$$${my}\:{attempt}\:{to}\:{improve}\:{upon} \\ $$$${the}\:{Cardano}'{s}\:{cubic}\:{formula}.. \\ $$$${or}\:{as}\:{an}\:{alternate}\:{to}\:{the} \\ $$$${trigonometric}\:{solution}... \\ $$$${here}\:{it}\:{goes}: \\ $$$$\:\:\:\:{X}^{\mathrm{3}} −{pX}−{q}=\mathrm{0} \\ $$$${let}\:\:\frac{{X}}{\:\sqrt{{p}}}\:=\:{x}\:;\:\:{and}\:{with}\:\frac{{q}}{{p}\sqrt{{p}}}\:=\:{c}\:, \\ $$$$\:\:\:\:{x}^{\mathrm{3}} −{x}−{c}=\mathrm{0} \\ $$$${let}\:\:\boldsymbol{{x}}=\left(\boldsymbol{{k}}+\mathrm{1}\right)\left(\boldsymbol{{p}}+\boldsymbol{{q}}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:=\left(\boldsymbol{{p}}+\boldsymbol{{kq}}\right)\:+\:\left(\boldsymbol{{kp}}+\boldsymbol{{q}}\right) \\ $$$$\:\:\:\:{p}^{\mathrm{3}} +{k}^{\mathrm{3}} {q}^{\mathrm{3}} +\mathrm{3}{kpq}\left({p}+{kq}\right)+ \\ $$$$\:\:\:\:{k}^{\mathrm{3}} {p}^{\mathrm{3}} +{q}^{\mathrm{3}} +\mathrm{3}{kpq}\left({kp}+{q}\right) \\ $$$$\:\:\:\:\:−\left({p}+{kq}\right)−\left({kp}+{q}\right)={c} \\ $$$$\Rightarrow \\ $$$$\:\:\:\left(\mathrm{1}+{k}^{\mathrm{3}} \right)\left({p}^{\mathrm{3}} +{q}^{\mathrm{3}} \right)+ \\ $$$$\:\:\:\left({p}+{kq}\right)\left(\mathrm{3}{kpq}−\mathrm{1}\right)+ \\ $$$$\:\:\:\left({kp}+{q}\right)\left(\mathrm{3}{kpq}−\mathrm{1}\right)=\:{c} \\ $$$${let}\:\:\:\mathrm{3}\boldsymbol{{kpq}}=\mathrm{1}\:\:\:\Rightarrow \\ $$$$\:\:\:{p}^{\mathrm{3}} +{q}^{\mathrm{3}} =\frac{{c}}{\mathrm{1}+{k}^{\mathrm{3}} } \\ $$$$\&\:\:\:{p}^{\mathrm{3}} {q}^{\mathrm{3}} =\frac{\mathrm{1}}{\mathrm{27}{k}^{\mathrm{3}} }\:\:\:;\:\:{hence} \\ $$$$\:\:\:{p}^{\mathrm{3}} ,\:{q}^{\mathrm{3}} \:\:=\: \\ $$$$\:\:\frac{{c}}{\mathrm{2}\left(\mathrm{1}+{k}^{\mathrm{3}} \right)}\pm\sqrt{\frac{{c}^{\mathrm{2}} }{\mathrm{4}\left(\mathrm{1}+{k}^{\mathrm{3}} \right)^{\mathrm{2}} }−\frac{\mathrm{1}}{\mathrm{27}{k}^{\mathrm{3}} }} \\ $$$${lets}\:{choose}\:{upon}\:{a}\:{value}\:{of}\:\boldsymbol{{k}} \\ $$$${such}\:{that}\:{D}=\mathrm{0} \\ $$$$\Rightarrow\:\:\mathrm{4}\left(\mathrm{1}+{k}^{\mathrm{3}} \right)^{\mathrm{2}} =\mathrm{27}{c}^{\mathrm{2}} {k}^{\mathrm{3}} \:\:\:.....\left({I}\right) \\ $$$$\:\:\left({just}\:{a}\:{quadratic}..\right) \\ $$$${first}\:{for}\:\:\:{c}^{\mathrm{2}} >\frac{\mathrm{8}}{\mathrm{27}}\:\:{we}\:\:{always}\:{can} \\ $$$${get}\:{two}\:{real}\:{k}\:{values},\:{and}\:{even} \\ $$$${p}={q}\:{then}. \\ $$$$\:\:{x}=\left({k}+\mathrm{1}\right)\left({p}+{q}\right) \\ $$$$\:\:\:\:=\:\mathrm{2}\left({k}+\mathrm{1}\right){p}\: \\ $$$$\:{x}=\mathrm{2}\left({k}+\mathrm{1}\right)\left[\frac{{c}}{\mathrm{2}\left(\mathrm{1}+{k}^{\mathrm{3}} \right)}\right]^{\mathrm{1}/\mathrm{3}} \\ $$$${but}\:{simply}\:\:{pq}=\frac{\mathrm{1}}{\mathrm{3}{k}}\:\:\:{hence} \\ $$$$\:{x}=\mathrm{2}\left({k}+\mathrm{1}\right){p}\:=\:\frac{\mathrm{2}\left({k}+\mathrm{1}\right)}{\:\sqrt{\mathrm{3}{k}}}\:\:. \\ $$$$\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_ \\ $$$${even}\:{for}\:\:{c}=\mathrm{1}\:\:{i}\:{dint}\:{get}\:{a} \\ $$$${correct}\:{answer},\:\:{please}\:{help} \\ $$$${error}-{freeing}\:{it}.\:\left(\mathcal{T}{hanks}!\right) \\ $$
Commented by MJS_new last updated on 04/Jan/21
$$\mathrm{after}\:\mathrm{substituting}\:{x}=\left({k}+\mathrm{1}\right)\left({p}+{q}\right)\:\mathrm{I}\:\mathrm{don}'\mathrm{t}\:\mathrm{get} \\ $$$$\mathrm{your}\:\mathrm{next}\:\mathrm{equation}. \\ $$$$ \\ $$$${p},\:{q}\:\mathrm{are}\:\mathrm{given}\:\Rightarrow\:\mathrm{from}\:\mathrm{the}\:\mathrm{moment}\:\mathrm{you}\:\mathrm{set} \\ $$$$\mathrm{3}{kpq}=\mathrm{1}\:\mathrm{also}\:{k}\:\mathrm{is}\:\mathrm{given}\:\Rightarrow\:{x}\:\mathrm{is}\:\mathrm{given} \\ $$
Commented by ajfour last updated on 04/Jan/21
$${because}\:{we}\:{put}\:\:\:\:\mathrm{3}{kpq}=\mathrm{1} \\ $$$${and}\:{use}\:{the}\:{equation},\:{so}\:{we}\:{get} \\ $$$${p}^{\mathrm{3}} +{q}^{\mathrm{3}} =\frac{{c}}{\mathrm{2}\left(\mathrm{1}+{k}^{\mathrm{3}} \right)} \\ $$$$\Rightarrow\:\:{if}\:{we}\:{give}\:{a}\:{value}\:{to}\:{k}\:{then} \\ $$$${x}=\left(\mathrm{1}+{k}\right)\left({p}+{q}\right)\:\:\:{is}\:{obtained} \\ $$$${in}\:{accordance}\:{with}\:{the}\:{eq}. \\ $$$$\:\:{x}^{\mathrm{3}} ={x}+{c}\:\:\blacksquare \\ $$