What method can i apply to solve this equation analytically?

Here is my equation $ $ 2{x^4} – 4x + 1 – ({x^3} – x – 1)\sqrt {3{x^3} + 1} = 0 % MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqipv0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaaeaaaaaaaaa8 % qacaaIYaGaamiEa8aadaahaaWcbeqaa8qacaaI0aaaaOGaeyOeI0Ia % aGinaiaadIhacqGHRaWkcaaIXaGaeyOeI0IaaiikaiaadIhapaWaaW % baaSqabeaapeGaaG4maaaakiabgkHiTiaadIhacqGHsislcaaIXaGa % aiykamaakaaapaqaa8qacaaIZaGaamiEa8aadaahaaWcbeqaa8qaca % aIZaaaaOGaey4kaSIaaGymaaWcbeaakiabg2da9iaaicdaaaa!4B9E! $ $ This equation has appeared on my advanced math test and all i have done was using MODE 7 on my calculator and it has given me a clue to prove this equation had no solution on $ $ \left[ { – \sqrt[3]{{\frac{1}{3}}};\frac{9}{5}} \right] % MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqipv0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaamWaaeaacq % GHsisldaGcbaqaamaalaaabaGaaGymaaqaaiaaiodaaaaaleaacaaI % ZaaaaOGaai4oamaalaaabaGaaGyoaaqaaiaaiwdaaaaacaGLBbGaay % zxaaaaaa!3DB0! $ $ If $ $ \left\{ \begin{array}{l}{x^3} – x – 1 > 0\x \in \left[ { – \sqrt[3]{{\frac{1}{3}}};\frac{9}{5}} \right]\end{array} \right. % MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqipv0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaiqaaqaabe % qaaiaadIhadaahaaWcbeqaaiaaiodaaaGccqGHsislcaWG4bGaeyOe % I0IaaGymaiabg6da+iaaicdaaeaacaWG4bGaeyicI48aamWaaeaacq % GHsisldaGcbaqaamaalaaabaGaaGymaaqaaiaaiodaaaaaleaacaaI % ZaaaaOGaai4oamaalaaabaGaaGyoaaqaaiaaiwdaaaaacaGLBbGaay % zxaaaaaiaawUhaaaaa!4897! $ $ so $ $ \begin{array}{l}\sqrt {3{x^3} + 1} \le \frac{{34\sqrt {10} }}{{25}} < \frac{9}{2}\f(x) = 2{x^4} – 4x + 1 – ({x^3} – x – 1)\sqrt {3{x^3} + 1} > 2{x^4} – 4x + 1 – \frac{9}{2}({x^3} – x – 1) = \frac{{4{x^4} – 9{x^3} + x + 11}}{2}\ = \frac{{4{{({x^2} – \frac{9}{8}x – \frac{9}{{10}})}^2} + \frac{{171}}{{80}}{x^2} – \frac{{71}}{{10}}x + \frac{{194}}{{25}}}}{2}\end{array} % MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqipv0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGceaqabeaadaGcaa % qaaiaaiodacaWG4bWaaWbaaSqabeaacaaIZaaaaOGaey4kaSIaaGym % aaWcbeaakiabgsMiJoaalaaabaGaaG4maiaaisdadaGcaaqaaiaaig % dacaaIWaaaleqaaaGcbaGaaGOmaiaaiwdaaaGaeyipaWZaaSaaaeaa % caaI5aaabaGaaGOmaaaaaeaacaWGMbGaaiikaiaadIhacaGGPaGaey % ypa0JaaGOmaiaadIhadaahaaWcbeqaaiaaisdaaaGccqGHsislcaaI % 0aGaamiEaiabgUcaRiaaigdacqGHsislcaGGOaGaamiEamaaCaaale % qabaGaaG4maaaakiabgkHiTiaadIhacqGHsislcaaIXaGaaiykamaa % kaaabaGaaG4maiaadIhadaahaaWcbeqaaiaaiodaaaGccqGHRaWkca % aIXaaaleqaaOGaeyOpa4JaaGOmaiaadIhadaahaaWcbeqaaiaaisda % aaGccqGHsislcaaI0aGaamiEaiabgUcaRiaaigdacqGHsisldaWcaa % qaaiaaiMdaaeaacaaIYaaaaiaacIcacaWG4bWaaWbaaSqabeaacaaI % ZaaaaOGaeyOeI0IaamiEaiabgkHiTiaaigdacaGGPaGaeyypa0ZaaS % aaaeaacaaI0aGaamiEamaaCaaaleqabaGaaGinaaaakiabgkHiTiaa % iMdacaWG4bWaaWbaaSqabeaacaaIZaaaaOGaey4kaSIaamiEaiabgU % caRiaaigdacaaIXaaabaGaaGOmaaaaaeaacqGH9aqpdaWcaaqaaiaa % isdacaGGOaGaamiEamaaCaaaleqabaGaaGOmaaaakiabgkHiTmaala % aabaGaaGyoaaqaaiaaiIdaaaGaamiEaiabgkHiTmaalaaabaGaaGyo % aaqaaiaaigdacaaIWaaaaiaacMcadaahaaWcbeqaaiaaikdaaaGccq % GHRaWkdaWcaaqaaiaaigdacaaI3aGaaGymaaqaaiaaiIdacaaIWaaa % aiaadIhadaahaaWcbeqaaiaaikdaaaGccqGHsisldaWcaaqaaiaaiE % dacaaIXaaabaGaaGymaiaaicdaaaGaamiEaiabgUcaRmaalaaabaGa % aGymaiaaiMdacaaI0aaabaGaaGOmaiaaiwdaaaaabaGaaGOmaaaaaa % aa!963F! $ $ >0

So this equation has no solution when $ $ \left\{ \begin{array}{l}{x^3} – x – 1 > 0\x \le \frac{9}{5}\end{array} \right. % MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqipv0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaiqaaqaabe % qaaiaadIhadaahaaWcbeqaaiaaiodaaaGccqGHsislcaWG4bGaeyOe % I0IaaGymaiabg6da+iaaicdaaeaacaWG4bGaeyizIm6aaSaaaeaaca % aI5aaabaGaaGynaaaaaaGaay5Eaaaaaa!42C0! $ $ I though this was an important thing to obtain. But i have no idea what to do further.

Mathematica cannot solve my simultaneous equations analytically but it can when I replace the variable with a number

I am attempting to solve three simultaneous equations for b,e and f in terms of another variable p. My code is:

Clear[b, e, f]  Solve[2/(1 - p) == -(2. b + e - f) && 2/(1 - p) == -2. e - p.b + 2. f - p.f && -2/(1 - p) == -2. f + p.b + 2. e - p.e, {b, e, f}] 

Which produces the error message: Solve::inex: Solve was unable to solve the system with inexact coefficients or the system obtained by direct rationalization of inexact numbers present in the system. Since many of the methods used by Solve require exact input, providing Solve with an exact version of the system may help.

However this is odd because I have solved these equations by hand and found a solution. Also when I change p to a number Mathematica can solve these equations easily which is correct. I need the solution in terms of p however. The reason I need this is to generalise to larger systems of equations even though I have this solution.

Does anyone know what the issue is? I am aware that there have been a few discussions on how to fix this error message in general, but that required numerical approaches to get the roots, which won’t give me the analytical solution I need. Interestingly also when I drop down to 2 simultaneous equations I can get the solution easily through Mathematica. Any help would be greatly appreciated. Thanks!

Why my pdf does not integrate to $1$ numerically even though it does analytically?

I’m running a simulation to verify my pdfs integrate to $ 1$ numerically. My problem is that pdfa outputs 1. but pdfg outputs 0.995075. I understand it’s almost $ 1$ but still this slight error affect my subsequent simulations. Is this error resulting from computations? Is it fixable?

Code

a = 9.6117; b = 0.1581;  lamdaA = 0.00001; ha = 100; pta = 1; etaaN = 3.2; etaaL = 3;  lamdaG = 0.001; hg = 60; ptg = 1; etag = 3.6;  PL[rr_] := 1/(1 + a Exp[b (a - 180/\[Pi] ArcTan[ha/rr])])  Eg[rr_] := (pta/ptg)^(2/etaaL) (rr^2 + hg^2)^(etag/etaaL) - ha^2 Ea[rr_] := (ptg/pta)^(2/etag) (rr^2 + ha^2)^(etaaL/etag) - hg^2  frg[rr_] := 2 \[Pi] lamdaG rr Exp[-\[Pi] lamdaG rr^2]; Frg[rr_] := 1 - Exp[-\[Pi] lamdaG rr^2];  fra[rr_] := 2 \[Pi] lamdaA rr Exp[-\[Pi] lamdaA rr^2]; Fra[rr_] := 1 - Exp[-\[Pi] lamdaA rr^2]  intt1[rr_] := PL[rr] fra[rr] ; intt2[rr_] := PL[rr] fra[rr] Frg[Sqrt[Ea[rr]]];  If [Eg[0] > 0 ,  ppa = ( NIntegrate[intt1[t] , {t, 0, \[Infinity]}] -       NIntegrate[intt2[t], {t, Sqrt[Eg[0]], \[Infinity]}]);,  ppa = (NIntegrate[intt2[t], {t, 0, \[Infinity]}]);]  fxa1[xx_] := (PL[xx] fra[xx])/ppa;  fxa2[xx_] := (PL[xx] fra[xx])/ppa (1 - Frg[Sqrt[Ea[xx]]]); fxg1[xx_] := (PL[xx] frg[xx])/(1 - ppa); fxg2[xx_] := ((PL[xx] frg[xx])/(1 - ppa)) (1 - Fra[Sqrt[Eg[xx]]]);  pdfa1 := NIntegrate[fxa1[t], {t, 0, Sqrt[Eg[0]]}] +     NIntegrate[fxa2[t], {t, Sqrt[Eg[0]], \[Infinity]}]; pdfa2 := NIntegrate[fxa2[t], {t, 0, \[Infinity]}];  If[(Eg[0] > 0 ),  pdfa = pdfa1,  pdfa = pdfa2]  pdfg1 := NIntegrate[fxg1[t], {t, 0, Sqrt[Ea[0]]}] +     NIntegrate[fxg2[t], {t, Sqrt[Ea[0]], \[Infinity]}]; pdfg2 := NIntegrate[fxg2[t], {t, 0, \[Infinity]}];  If[(Ea[0] > 0 ),  pdfg = pdfg1,  pdfg = pdfg2]