How to solve $x^6- 4x^3+4=0$?












1












$begingroup$


I'm working through an exercise with the question:




Use substitution to solve for $x$ in the following equation
$x^6 - 4x^3 + 4 = 0$




The solution given is $sqrt[3]2$ however there is no working in between shown and I'm at a loss for how to arrive at this.



I started trying to factor our the polynomials $x^6$ and $4x^3$ but that didn't really get me anywhere:



$x^3(x^3 - 4) = -4$ is as far as I got.



How can I arrive at $sqrt[3]2$ with each step shown in between?






algebra-precalculus polynomials roots







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  • 5




    $begingroup$
    Let $u=x^3$. Then your equation will turn into a quadratic equation.
    $endgroup$
    – user9077
    Dec 11 '18 at 17:07
















1












$begingroup$


I'm working through an exercise with the question:




Use substitution to solve for $x$ in the following equation
$x^6 - 4x^3 + 4 = 0$




The solution given is $sqrt[3]2$ however there is no working in between shown and I'm at a loss for how to arrive at this.



I started trying to factor our the polynomials $x^6$ and $4x^3$ but that didn't really get me anywhere:



$x^3(x^3 - 4) = -4$ is as far as I got.



How can I arrive at $sqrt[3]2$ with each step shown in between?






algebra-precalculus polynomials roots







share|cite|improve this question











$endgroup$








  • 5




    $begingroup$
    Let $u=x^3$. Then your equation will turn into a quadratic equation.
    $endgroup$
    – user9077
    Dec 11 '18 at 17:07














1












1








1





$begingroup$


I'm working through an exercise with the question:




Use substitution to solve for $x$ in the following equation
$x^6 - 4x^3 + 4 = 0$




The solution given is $sqrt[3]2$ however there is no working in between shown and I'm at a loss for how to arrive at this.



I started trying to factor our the polynomials $x^6$ and $4x^3$ but that didn't really get me anywhere:



$x^3(x^3 - 4) = -4$ is as far as I got.



How can I arrive at $sqrt[3]2$ with each step shown in between?






algebra-precalculus polynomials roots







share|cite|improve this question











$endgroup$




I'm working through an exercise with the question:




Use substitution to solve for $x$ in the following equation
$x^6 - 4x^3 + 4 = 0$




The solution given is $sqrt[3]2$ however there is no working in between shown and I'm at a loss for how to arrive at this.



I started trying to factor our the polynomials $x^6$ and $4x^3$ but that didn't really get me anywhere:



$x^3(x^3 - 4) = -4$ is as far as I got.



How can I arrive at $sqrt[3]2$ with each step shown in between?






algebra-precalculus polynomials roots




algebra-precalculus polynomials roots






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 11 '18 at 17:19









Brahadeesh

6,32942363




6,32942363










asked Dec 11 '18 at 17:04









Doug FirDoug Fir

3227




3227








  • 5




    $begingroup$
    Let $u=x^3$. Then your equation will turn into a quadratic equation.
    $endgroup$
    – user9077
    Dec 11 '18 at 17:07














  • 5




    $begingroup$
    Let $u=x^3$. Then your equation will turn into a quadratic equation.
    $endgroup$
    – user9077
    Dec 11 '18 at 17:07








5




5




$begingroup$
Let $u=x^3$. Then your equation will turn into a quadratic equation.
$endgroup$
– user9077
Dec 11 '18 at 17:07




$begingroup$
Let $u=x^3$. Then your equation will turn into a quadratic equation.
$endgroup$
– user9077
Dec 11 '18 at 17:07










3 Answers
3






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7












$begingroup$

Hint: put $y = x^3$, and you get a quadratic equation.






share|cite|improve this answer









$endgroup$





















    3












    $begingroup$

    Use that $x^6-4x^3+4=(x^3-2)^2$






    share|cite|improve this answer









    $endgroup$





















      3












      $begingroup$

      A simple substitution gets the job done. Let $t = x^3$.



      $$x^3(x^3-4) = -4 iff t(t-4) = -4 iff t = 2$$



      $$t = 2 iff x^3 = 2 iff x = sqrt[3]{2}$$



      As a more general tip, whenever you have “quadratic-like” equations, make a substitution to reach a normal quadratic.



      $$ax^{2n}+bx^n+c = 0$$



      Here, letting $t = x^n$ reduces the problem to solving for $t$ in $at^2+bt+c = 0$.






      share|cite|improve this answer











      $endgroup$













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        3 Answers
        3






        active

        oldest

        votes








        3 Answers
        3






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes









        7












        $begingroup$

        Hint: put $y = x^3$, and you get a quadratic equation.






        share|cite|improve this answer









        $endgroup$


















          7












          $begingroup$

          Hint: put $y = x^3$, and you get a quadratic equation.






          share|cite|improve this answer









          $endgroup$
















            7












            7








            7





            $begingroup$

            Hint: put $y = x^3$, and you get a quadratic equation.






            share|cite|improve this answer









            $endgroup$



            Hint: put $y = x^3$, and you get a quadratic equation.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Dec 11 '18 at 17:07









            DeepSeaDeepSea

            71.2k54487




            71.2k54487























                3












                $begingroup$

                Use that $x^6-4x^3+4=(x^3-2)^2$






                share|cite|improve this answer









                $endgroup$


















                  3












                  $begingroup$

                  Use that $x^6-4x^3+4=(x^3-2)^2$






                  share|cite|improve this answer









                  $endgroup$
















                    3












                    3








                    3





                    $begingroup$

                    Use that $x^6-4x^3+4=(x^3-2)^2$






                    share|cite|improve this answer









                    $endgroup$



                    Use that $x^6-4x^3+4=(x^3-2)^2$







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Dec 11 '18 at 17:12









                    VasyaVasya

                    3,3271515




                    3,3271515























                        3












                        $begingroup$

                        A simple substitution gets the job done. Let $t = x^3$.



                        $$x^3(x^3-4) = -4 iff t(t-4) = -4 iff t = 2$$



                        $$t = 2 iff x^3 = 2 iff x = sqrt[3]{2}$$



                        As a more general tip, whenever you have “quadratic-like” equations, make a substitution to reach a normal quadratic.



                        $$ax^{2n}+bx^n+c = 0$$



                        Here, letting $t = x^n$ reduces the problem to solving for $t$ in $at^2+bt+c = 0$.






                        share|cite|improve this answer











                        $endgroup$


















                          3












                          $begingroup$

                          A simple substitution gets the job done. Let $t = x^3$.



                          $$x^3(x^3-4) = -4 iff t(t-4) = -4 iff t = 2$$



                          $$t = 2 iff x^3 = 2 iff x = sqrt[3]{2}$$



                          As a more general tip, whenever you have “quadratic-like” equations, make a substitution to reach a normal quadratic.



                          $$ax^{2n}+bx^n+c = 0$$



                          Here, letting $t = x^n$ reduces the problem to solving for $t$ in $at^2+bt+c = 0$.






                          share|cite|improve this answer











                          $endgroup$
















                            3












                            3








                            3





                            $begingroup$

                            A simple substitution gets the job done. Let $t = x^3$.



                            $$x^3(x^3-4) = -4 iff t(t-4) = -4 iff t = 2$$



                            $$t = 2 iff x^3 = 2 iff x = sqrt[3]{2}$$



                            As a more general tip, whenever you have “quadratic-like” equations, make a substitution to reach a normal quadratic.



                            $$ax^{2n}+bx^n+c = 0$$



                            Here, letting $t = x^n$ reduces the problem to solving for $t$ in $at^2+bt+c = 0$.






                            share|cite|improve this answer











                            $endgroup$



                            A simple substitution gets the job done. Let $t = x^3$.



                            $$x^3(x^3-4) = -4 iff t(t-4) = -4 iff t = 2$$



                            $$t = 2 iff x^3 = 2 iff x = sqrt[3]{2}$$



                            As a more general tip, whenever you have “quadratic-like” equations, make a substitution to reach a normal quadratic.



                            $$ax^{2n}+bx^n+c = 0$$



                            Here, letting $t = x^n$ reduces the problem to solving for $t$ in $at^2+bt+c = 0$.







                            share|cite|improve this answer














                            share|cite|improve this answer



                            share|cite|improve this answer








                            edited Dec 11 '18 at 17:21

























                            answered Dec 11 '18 at 17:15









                            KM101KM101

                            5,9251524




                            5,9251524






























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