Real solution of x3+1=xx^3 +1=x

I was just wondering about its different solutions while one which I did was Cardan's solution

x3x^3 -xx +1=0

Let xx =y+zy + z and yzyz =13\frac {1}{3} \Rightarrow 3yz13yz - 1 =0

x3x^3 =(y+z)3(y+z)^3 =y3+z3+3yz(y+z)y^3 + z^3 + 3yz (y+z) =y3+z3+3yzxy^3 + z^3 +3yzx

Putting in the equation value of x3x^3

y3+z3+(3yz1)x+1y^3 + z^3 + (3yz-1)x +1

y3+z3+1y^3 + z^3 +1 \Rightarrow y3+z3y^3+z^3 =-1 and also yzyz =13\frac {1}{3} \Rightarrow y3z3y^3z^3 =127\frac {1}{27}

Making quadratic in y3y^3 and z3z^3 solving we can get y and z and thus y+z=x by making it as factor we can get other solution which are complex.

Can you provide such simillar solution and what if question was x4x^4 +1=xx

Note by Tarun Garg
6 years ago

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8 votes

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The simplest way to solve this quartic, without wheeling in a load of theory, is to try to complete the square. We can write x4x+1  =  (x2+a)2b(x+c)2 x^4 - x + 1 \; = \; (x^2 + a)^2 - b(x + c)^2 provided that b=2ab=2a, 2bc=12bc = 1 and a2bc2=1a^2 - bc^2 = 1. Then b=2ab = 2a, c=14ac = \tfrac{1}{4a}, and aa satisfies the equation a218a  =  1 a^2 - \tfrac{1}{8a} \; = \; 1 Thus we need to solve a cubic equation to find aa. Once we have done this, we can use aa, bb and cc to factorise x4x+1x^4 - x + 1 as a product of two quadratics, and then solving the equation x4x+1=0x^4 - x + 1 = 0 is straightforward.

It is possible to solve all quartics with no x3x^3 term this way; completing the square aims to write the quartic as a difference of two squares, and the condition that needs to be solved to make this possible requires us to solve a cubic equation. If we can solve cubics, we can solve quartics. The restriction that the quartic should have no x3x^3 term is no problem; any cubic can be converted into one without an x3x^3 term simply by translation, regarding it as a function of x+ux+u for a suitable uu.

Mark Hennings - 6 years ago

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A more theoretically elegant way to solve the quartic is the following. If t1,t2,t3,t4t_1,t_2,t_3,t_4 are the roots of the quartic x4+ax3+bc2+cx+d=0 x^4 + ax^3 + bc^2 + cx + d = 0 then consider s1=t1t2+t3t4s2=t1t3+t2t4s3=t1t4+t2t3 s_1 = t_1t_2+ t_3t_4 \qquad s_2 = t_1t_3+t_2t_4 \qquad s_3 = t_1t_4+t_2t_3 We can use the usual polynomial root techniques to identify s1+s2+s3(=b)s_1+s_2+s_3 (= b), s1s2+s1s3+s2s3s_1s_2+s_1s_3+s_2s_3 and s1s2s3s_1s_2s_3 in terms of a,b,c,da,b,c,d, and hence find a cubic equation whose roots are s1,s2,s3s_1,s_2,s_3. Once we know the values of s1,s2,s3s_1,s_2,s_3, finding the values of t1,t2,t3,t4t_1,t_2,t_3,t_4 is not difficult.

Mark Hennings - 6 years ago

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That's a really good method getting lots of things to think here.

Tarun Garg - 6 years ago

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Real solution

x=(23233212)13+13(23233212)13x={\left( \frac{\sqrt{23}}{2\,{3}^{\frac{3}{2}}}-\frac{1}{2}\right) }^{\frac{1}{3}}+\frac{1}{3\,{\left( \frac{\sqrt{23}}{2\,{3}^{\frac{3}{2}}}-\frac{1}{2}\right) }^{\frac{1}{3}}}

Joseph Gomes - 6 years ago

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any process for solution of x^5+1=x

ankan ghosh - 6 years ago

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No. The quintic polynomial equation x5x+1=0x^5 - x +1 = 0 cannot be solved by these techniques (adding, subtracting, multiplying, dividing, taking nnth roots). Collectively, these techniques are called "using radicals". While we can solve quadratics, cubics and quartics by radicals, we cannot (in general) solve quintics by radicals. This result is one of the great triumphs of Galois Theory, and relies on the fact that the permutation group S5S_5 of five symbols is a much more complicated group (in a particular sense) than the permutation group S4S_4 of four symbols.

Mark Hennings - 6 years ago

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