Roots for a cubic equations












0












$begingroup$


Find all real or complex solutions of the simultaneous equations
$$begin{aligned}x+y+z&=3\x^2+y^2+z^2&=3\x^3+y^3+z^3&=end{aligned}$$
I assumed that $x,y,z$ are 3 roots for a cubic equation such that $$W^3+aW^2+bW+c=0$$
and using relationship between roots of a polynomial equation I got $a=-3,; b=3,; c=-1.$

Which means the normal cubic equation is $（W-1)^3.$

My question is: are there possible complex solutions for such the relationship between $x,y$ and $z$ still be true. And please help me find the complex solutions as well.

Thank you






polynomials complex-numbers







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$endgroup$












  • $begingroup$
    what is with the third term?
    $endgroup$
    – Dr. Sonnhard Graubner
    Sep 27 '15 at 18:56






  • 2




    $begingroup$
    The cubic equation is incomplete.
    $endgroup$
    – Bernard
    Sep 27 '15 at 18:57
















0












$begingroup$


Find all real or complex solutions of the simultaneous equations
$$begin{aligned}x+y+z&=3\x^2+y^2+z^2&=3\x^3+y^3+z^3&=end{aligned}$$
I assumed that $x,y,z$ are 3 roots for a cubic equation such that $$W^3+aW^2+bW+c=0$$
and using relationship between roots of a polynomial equation I got $a=-3,; b=3,; c=-1.$

Which means the normal cubic equation is $（W-1)^3.$

My question is: are there possible complex solutions for such the relationship between $x,y$ and $z$ still be true. And please help me find the complex solutions as well.

Thank you






polynomials complex-numbers







share|cite|improve this question











$endgroup$












  • $begingroup$
    what is with the third term?
    $endgroup$
    – Dr. Sonnhard Graubner
    Sep 27 '15 at 18:56






  • 2




    $begingroup$
    The cubic equation is incomplete.
    $endgroup$
    – Bernard
    Sep 27 '15 at 18:57














0












0








0





$begingroup$


Find all real or complex solutions of the simultaneous equations
$$begin{aligned}x+y+z&=3\x^2+y^2+z^2&=3\x^3+y^3+z^3&=end{aligned}$$
I assumed that $x,y,z$ are 3 roots for a cubic equation such that $$W^3+aW^2+bW+c=0$$
and using relationship between roots of a polynomial equation I got $a=-3,; b=3,; c=-1.$

Which means the normal cubic equation is $（W-1)^3.$

My question is: are there possible complex solutions for such the relationship between $x,y$ and $z$ still be true. And please help me find the complex solutions as well.

Thank you






polynomials complex-numbers







share|cite|improve this question











$endgroup$




Find all real or complex solutions of the simultaneous equations
$$begin{aligned}x+y+z&=3\x^2+y^2+z^2&=3\x^3+y^3+z^3&=end{aligned}$$
I assumed that $x,y,z$ are 3 roots for a cubic equation such that $$W^3+aW^2+bW+c=0$$
and using relationship between roots of a polynomial equation I got $a=-3,; b=3,; c=-1.$

Which means the normal cubic equation is $（W-1)^3.$

My question is: are there possible complex solutions for such the relationship between $x,y$ and $z$ still be true. And please help me find the complex solutions as well.

Thank you






polynomials complex-numbers




polynomials complex-numbers






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 30 '18 at 20:54









user376343

3,9934829




3,9934829










asked Sep 27 '15 at 18:52









Joseph ZhuJoseph Zhu

112




112












  • $begingroup$
    what is with the third term?
    $endgroup$
    – Dr. Sonnhard Graubner
    Sep 27 '15 at 18:56






  • 2




    $begingroup$
    The cubic equation is incomplete.
    $endgroup$
    – Bernard
    Sep 27 '15 at 18:57


















  • $begingroup$
    what is with the third term?
    $endgroup$
    – Dr. Sonnhard Graubner
    Sep 27 '15 at 18:56






  • 2




    $begingroup$
    The cubic equation is incomplete.
    $endgroup$
    – Bernard
    Sep 27 '15 at 18:57
















$begingroup$
what is with the third term?
$endgroup$
– Dr. Sonnhard Graubner
Sep 27 '15 at 18:56




$begingroup$
what is with the third term?
$endgroup$
– Dr. Sonnhard Graubner
Sep 27 '15 at 18:56




2




2




$begingroup$
The cubic equation is incomplete.
$endgroup$
– Bernard
Sep 27 '15 at 18:57




$begingroup$
The cubic equation is incomplete.
$endgroup$
– Bernard
Sep 27 '15 at 18:57










3 Answers
3






active

oldest

votes


















0












$begingroup$

The cubic equation is incomplete - I assume that you left out equal to $3$ on the RHS in the last equation. In that case it is easy to see that the only solution is given by $(x,y,z)=(1,1,1)$. If you meant the equations $x+y+z=x^2+y^2+z^2=3$ and $x^3+y^3+z^3=0$, then there are more interesting solutions, i.e., $x=frac{sqrt{-3}+3}{2}$, $y=frac{-sqrt{-3}+3}{2}$ and $z=0$.






share|cite|improve this answer











$endgroup$





















    0












    $begingroup$

    $textbf{Hint: }$ Try the method of Newton's Sums.$$$$
    Let $$P_i=x_1^i+x_2^i+x_3^i$$



    Assuming that the given equations are $$x_1+x_2+x_3=3, x_1^2+_2^2+x_3^2=3, x_1^3+x_2^3+x_3^3=k,$$



    Let $f(x)=ax^3+bx^2+cx+d$ have roots $x_1,x_2,x_3$$$$$ We want to find the values of $a,b,c,d$ so as to find the roots of $f(x)$. This will furnish us with the values of $x_1,x_2,x_3$. $$$$
    Also, $$ e_1 = sum x_i = - frac{b}{a} $$
    $$e_2 = sum x_i x_j = frac{ c } { a} $$
    $$e_3 = sum x_i x_j x_k = - frac{ d } { a}
    $$ $$$$
    Lastly, the recurrence relation for $P_i$ for a polynomial of degree $k$ is as follows:$$$$



    For $ i leq k - 1 $, we have the formulas



    $$P_1 = e_1 times 1 $$
    $$P_2 = e_1 P_1 - e_2 times 2 $$
    $$P_3 = e_1 P_2 - e_2 P_1 + e_3 times 3 $$
    $$$$



    Since $P_1=3, P_2=3, P_3=k,$ we can calculate the corresponding values of $e_1, e_2, e_3$. This will give us the values of the coefficients $a, b, c,d$ of $f(x)$. It then becomes a simple task of factorizing $f(x)$ to find the values of its roots.






    share|cite|improve this answer











    $endgroup$





















      0












      $begingroup$

      Cauchy Schwarz Inequality



      $$(x^2+y^2+z^2)(1^2+1^2+1^2)geq (x+y+z)^2$$



      equality hold when $displaystyle x=y=z$



      So we have $(x,y,z)=(1,1,1)$.






      share|cite|improve this answer









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        3 Answers
        3






        active

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        3 Answers
        3






        active

        oldest

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        active

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        active

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        0












        $begingroup$

        The cubic equation is incomplete - I assume that you left out equal to $3$ on the RHS in the last equation. In that case it is easy to see that the only solution is given by $(x,y,z)=(1,1,1)$. If you meant the equations $x+y+z=x^2+y^2+z^2=3$ and $x^3+y^3+z^3=0$, then there are more interesting solutions, i.e., $x=frac{sqrt{-3}+3}{2}$, $y=frac{-sqrt{-3}+3}{2}$ and $z=0$.






        share|cite|improve this answer











        $endgroup$


















          0












          $begingroup$

          The cubic equation is incomplete - I assume that you left out equal to $3$ on the RHS in the last equation. In that case it is easy to see that the only solution is given by $(x,y,z)=(1,1,1)$. If you meant the equations $x+y+z=x^2+y^2+z^2=3$ and $x^3+y^3+z^3=0$, then there are more interesting solutions, i.e., $x=frac{sqrt{-3}+3}{2}$, $y=frac{-sqrt{-3}+3}{2}$ and $z=0$.






          share|cite|improve this answer











          $endgroup$
















            0












            0








            0





            $begingroup$

            The cubic equation is incomplete - I assume that you left out equal to $3$ on the RHS in the last equation. In that case it is easy to see that the only solution is given by $(x,y,z)=(1,1,1)$. If you meant the equations $x+y+z=x^2+y^2+z^2=3$ and $x^3+y^3+z^3=0$, then there are more interesting solutions, i.e., $x=frac{sqrt{-3}+3}{2}$, $y=frac{-sqrt{-3}+3}{2}$ and $z=0$.






            share|cite|improve this answer











            $endgroup$



            The cubic equation is incomplete - I assume that you left out equal to $3$ on the RHS in the last equation. In that case it is easy to see that the only solution is given by $(x,y,z)=(1,1,1)$. If you meant the equations $x+y+z=x^2+y^2+z^2=3$ and $x^3+y^3+z^3=0$, then there are more interesting solutions, i.e., $x=frac{sqrt{-3}+3}{2}$, $y=frac{-sqrt{-3}+3}{2}$ and $z=0$.







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Sep 28 '15 at 17:48

























            answered Sep 27 '15 at 18:57









            Dietrich BurdeDietrich Burde

            82.8k649107




            82.8k649107























                0












                $begingroup$

                $textbf{Hint: }$ Try the method of Newton's Sums.$$$$
                Let $$P_i=x_1^i+x_2^i+x_3^i$$



                Assuming that the given equations are $$x_1+x_2+x_3=3, x_1^2+_2^2+x_3^2=3, x_1^3+x_2^3+x_3^3=k,$$



                Let $f(x)=ax^3+bx^2+cx+d$ have roots $x_1,x_2,x_3$$$$$ We want to find the values of $a,b,c,d$ so as to find the roots of $f(x)$. This will furnish us with the values of $x_1,x_2,x_3$. $$$$
                Also, $$ e_1 = sum x_i = - frac{b}{a} $$
                $$e_2 = sum x_i x_j = frac{ c } { a} $$
                $$e_3 = sum x_i x_j x_k = - frac{ d } { a}
                $$ $$$$
                Lastly, the recurrence relation for $P_i$ for a polynomial of degree $k$ is as follows:$$$$



                For $ i leq k - 1 $, we have the formulas



                $$P_1 = e_1 times 1 $$
                $$P_2 = e_1 P_1 - e_2 times 2 $$
                $$P_3 = e_1 P_2 - e_2 P_1 + e_3 times 3 $$
                $$$$



                Since $P_1=3, P_2=3, P_3=k,$ we can calculate the corresponding values of $e_1, e_2, e_3$. This will give us the values of the coefficients $a, b, c,d$ of $f(x)$. It then becomes a simple task of factorizing $f(x)$ to find the values of its roots.






                share|cite|improve this answer











                $endgroup$


















                  0












                  $begingroup$

                  $textbf{Hint: }$ Try the method of Newton's Sums.$$$$
                  Let $$P_i=x_1^i+x_2^i+x_3^i$$



                  Assuming that the given equations are $$x_1+x_2+x_3=3, x_1^2+_2^2+x_3^2=3, x_1^3+x_2^3+x_3^3=k,$$



                  Let $f(x)=ax^3+bx^2+cx+d$ have roots $x_1,x_2,x_3$$$$$ We want to find the values of $a,b,c,d$ so as to find the roots of $f(x)$. This will furnish us with the values of $x_1,x_2,x_3$. $$$$
                  Also, $$ e_1 = sum x_i = - frac{b}{a} $$
                  $$e_2 = sum x_i x_j = frac{ c } { a} $$
                  $$e_3 = sum x_i x_j x_k = - frac{ d } { a}
                  $$ $$$$
                  Lastly, the recurrence relation for $P_i$ for a polynomial of degree $k$ is as follows:$$$$



                  For $ i leq k - 1 $, we have the formulas



                  $$P_1 = e_1 times 1 $$
                  $$P_2 = e_1 P_1 - e_2 times 2 $$
                  $$P_3 = e_1 P_2 - e_2 P_1 + e_3 times 3 $$
                  $$$$



                  Since $P_1=3, P_2=3, P_3=k,$ we can calculate the corresponding values of $e_1, e_2, e_3$. This will give us the values of the coefficients $a, b, c,d$ of $f(x)$. It then becomes a simple task of factorizing $f(x)$ to find the values of its roots.






                  share|cite|improve this answer











                  $endgroup$
















                    0












                    0








                    0





                    $begingroup$

                    $textbf{Hint: }$ Try the method of Newton's Sums.$$$$
                    Let $$P_i=x_1^i+x_2^i+x_3^i$$



                    Assuming that the given equations are $$x_1+x_2+x_3=3, x_1^2+_2^2+x_3^2=3, x_1^3+x_2^3+x_3^3=k,$$



                    Let $f(x)=ax^3+bx^2+cx+d$ have roots $x_1,x_2,x_3$$$$$ We want to find the values of $a,b,c,d$ so as to find the roots of $f(x)$. This will furnish us with the values of $x_1,x_2,x_3$. $$$$
                    Also, $$ e_1 = sum x_i = - frac{b}{a} $$
                    $$e_2 = sum x_i x_j = frac{ c } { a} $$
                    $$e_3 = sum x_i x_j x_k = - frac{ d } { a}
                    $$ $$$$
                    Lastly, the recurrence relation for $P_i$ for a polynomial of degree $k$ is as follows:$$$$



                    For $ i leq k - 1 $, we have the formulas



                    $$P_1 = e_1 times 1 $$
                    $$P_2 = e_1 P_1 - e_2 times 2 $$
                    $$P_3 = e_1 P_2 - e_2 P_1 + e_3 times 3 $$
                    $$$$



                    Since $P_1=3, P_2=3, P_3=k,$ we can calculate the corresponding values of $e_1, e_2, e_3$. This will give us the values of the coefficients $a, b, c,d$ of $f(x)$. It then becomes a simple task of factorizing $f(x)$ to find the values of its roots.






                    share|cite|improve this answer











                    $endgroup$



                    $textbf{Hint: }$ Try the method of Newton's Sums.$$$$
                    Let $$P_i=x_1^i+x_2^i+x_3^i$$



                    Assuming that the given equations are $$x_1+x_2+x_3=3, x_1^2+_2^2+x_3^2=3, x_1^3+x_2^3+x_3^3=k,$$



                    Let $f(x)=ax^3+bx^2+cx+d$ have roots $x_1,x_2,x_3$$$$$ We want to find the values of $a,b,c,d$ so as to find the roots of $f(x)$. This will furnish us with the values of $x_1,x_2,x_3$. $$$$
                    Also, $$ e_1 = sum x_i = - frac{b}{a} $$
                    $$e_2 = sum x_i x_j = frac{ c } { a} $$
                    $$e_3 = sum x_i x_j x_k = - frac{ d } { a}
                    $$ $$$$
                    Lastly, the recurrence relation for $P_i$ for a polynomial of degree $k$ is as follows:$$$$



                    For $ i leq k - 1 $, we have the formulas



                    $$P_1 = e_1 times 1 $$
                    $$P_2 = e_1 P_1 - e_2 times 2 $$
                    $$P_3 = e_1 P_2 - e_2 P_1 + e_3 times 3 $$
                    $$$$



                    Since $P_1=3, P_2=3, P_3=k,$ we can calculate the corresponding values of $e_1, e_2, e_3$. This will give us the values of the coefficients $a, b, c,d$ of $f(x)$. It then becomes a simple task of factorizing $f(x)$ to find the values of its roots.







                    share|cite|improve this answer














                    share|cite|improve this answer



                    share|cite|improve this answer








                    edited Jun 2 '18 at 8:00

























                    answered Jun 2 '18 at 7:50









                    IshanIshan

                    2,0221927




                    2,0221927























                        0












                        $begingroup$

                        Cauchy Schwarz Inequality



                        $$(x^2+y^2+z^2)(1^2+1^2+1^2)geq (x+y+z)^2$$



                        equality hold when $displaystyle x=y=z$



                        So we have $(x,y,z)=(1,1,1)$.






                        share|cite|improve this answer









                        $endgroup$


















                          0












                          $begingroup$

                          Cauchy Schwarz Inequality



                          $$(x^2+y^2+z^2)(1^2+1^2+1^2)geq (x+y+z)^2$$



                          equality hold when $displaystyle x=y=z$



                          So we have $(x,y,z)=(1,1,1)$.






                          share|cite|improve this answer









                          $endgroup$
















                            0












                            0








                            0





                            $begingroup$

                            Cauchy Schwarz Inequality



                            $$(x^2+y^2+z^2)(1^2+1^2+1^2)geq (x+y+z)^2$$



                            equality hold when $displaystyle x=y=z$



                            So we have $(x,y,z)=(1,1,1)$.






                            share|cite|improve this answer









                            $endgroup$



                            Cauchy Schwarz Inequality



                            $$(x^2+y^2+z^2)(1^2+1^2+1^2)geq (x+y+z)^2$$



                            equality hold when $displaystyle x=y=z$



                            So we have $(x,y,z)=(1,1,1)$.







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Jun 2 '18 at 8:04









                            DXTDXT

                            5,8872733




                            5,8872733






























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