How to expand $x^{1/3}-c^{1/3}$ into $(x-c)y$ for some $y$












0












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How to expand $x^{1/3}-c^{1/3}$ into $(x-c)y$ for some $y$




I know $x^3-c^3=(x-c)(x^2+xc+c^2)$ but I can't figure out how to pull this off with $1/3$ instead of $3$






factoring binomial-theorem







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$endgroup$












  • $begingroup$
    I don't think any similar rule applies for non-integers powers in terms of factorising polynomials
    $endgroup$
    – Henry Lee
    Dec 5 '18 at 1:13










  • $begingroup$
    Technically speaking that isn't "expanding" so much as factoring.
    $endgroup$
    – fleablood
    Dec 5 '18 at 1:28






  • 1




    $begingroup$
    "I don't think any similar rule applies for non-integers powers in terms of factorising polynomials" Sure there are. Just take $sqrt[k] x = M$ and $x = M^k$ anc $sqrt[k]c = N$. Then $x -c = (M^k - N^k)(M^{k-1}+ ... N^{k-1})= (sqrt[k] x - sqrt[k] c)(sqrt[k] x^{k-1} +.... + sqrt[k]c^{k-1})$
    $endgroup$
    – fleablood
    Dec 5 '18 at 1:32
















0












$begingroup$



How to expand $x^{1/3}-c^{1/3}$ into $(x-c)y$ for some $y$




I know $x^3-c^3=(x-c)(x^2+xc+c^2)$ but I can't figure out how to pull this off with $1/3$ instead of $3$






factoring binomial-theorem







share|cite|improve this question











$endgroup$












  • $begingroup$
    I don't think any similar rule applies for non-integers powers in terms of factorising polynomials
    $endgroup$
    – Henry Lee
    Dec 5 '18 at 1:13










  • $begingroup$
    Technically speaking that isn't "expanding" so much as factoring.
    $endgroup$
    – fleablood
    Dec 5 '18 at 1:28






  • 1




    $begingroup$
    "I don't think any similar rule applies for non-integers powers in terms of factorising polynomials" Sure there are. Just take $sqrt[k] x = M$ and $x = M^k$ anc $sqrt[k]c = N$. Then $x -c = (M^k - N^k)(M^{k-1}+ ... N^{k-1})= (sqrt[k] x - sqrt[k] c)(sqrt[k] x^{k-1} +.... + sqrt[k]c^{k-1})$
    $endgroup$
    – fleablood
    Dec 5 '18 at 1:32














0












0








0





$begingroup$



How to expand $x^{1/3}-c^{1/3}$ into $(x-c)y$ for some $y$




I know $x^3-c^3=(x-c)(x^2+xc+c^2)$ but I can't figure out how to pull this off with $1/3$ instead of $3$






factoring binomial-theorem







share|cite|improve this question











$endgroup$





How to expand $x^{1/3}-c^{1/3}$ into $(x-c)y$ for some $y$




I know $x^3-c^3=(x-c)(x^2+xc+c^2)$ but I can't figure out how to pull this off with $1/3$ instead of $3$






factoring binomial-theorem




factoring binomial-theorem






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 5 '18 at 15:05









Did

247k23222458




247k23222458










asked Dec 5 '18 at 1:11









Albert DiazAlbert Diaz

925




925












  • $begingroup$
    I don't think any similar rule applies for non-integers powers in terms of factorising polynomials
    $endgroup$
    – Henry Lee
    Dec 5 '18 at 1:13










  • $begingroup$
    Technically speaking that isn't "expanding" so much as factoring.
    $endgroup$
    – fleablood
    Dec 5 '18 at 1:28






  • 1




    $begingroup$
    "I don't think any similar rule applies for non-integers powers in terms of factorising polynomials" Sure there are. Just take $sqrt[k] x = M$ and $x = M^k$ anc $sqrt[k]c = N$. Then $x -c = (M^k - N^k)(M^{k-1}+ ... N^{k-1})= (sqrt[k] x - sqrt[k] c)(sqrt[k] x^{k-1} +.... + sqrt[k]c^{k-1})$
    $endgroup$
    – fleablood
    Dec 5 '18 at 1:32


















  • $begingroup$
    I don't think any similar rule applies for non-integers powers in terms of factorising polynomials
    $endgroup$
    – Henry Lee
    Dec 5 '18 at 1:13










  • $begingroup$
    Technically speaking that isn't "expanding" so much as factoring.
    $endgroup$
    – fleablood
    Dec 5 '18 at 1:28






  • 1




    $begingroup$
    "I don't think any similar rule applies for non-integers powers in terms of factorising polynomials" Sure there are. Just take $sqrt[k] x = M$ and $x = M^k$ anc $sqrt[k]c = N$. Then $x -c = (M^k - N^k)(M^{k-1}+ ... N^{k-1})= (sqrt[k] x - sqrt[k] c)(sqrt[k] x^{k-1} +.... + sqrt[k]c^{k-1})$
    $endgroup$
    – fleablood
    Dec 5 '18 at 1:32
















$begingroup$
I don't think any similar rule applies for non-integers powers in terms of factorising polynomials
$endgroup$
– Henry Lee
Dec 5 '18 at 1:13




$begingroup$
I don't think any similar rule applies for non-integers powers in terms of factorising polynomials
$endgroup$
– Henry Lee
Dec 5 '18 at 1:13












$begingroup$
Technically speaking that isn't "expanding" so much as factoring.
$endgroup$
– fleablood
Dec 5 '18 at 1:28




$begingroup$
Technically speaking that isn't "expanding" so much as factoring.
$endgroup$
– fleablood
Dec 5 '18 at 1:28




1




1




$begingroup$
"I don't think any similar rule applies for non-integers powers in terms of factorising polynomials" Sure there are. Just take $sqrt[k] x = M$ and $x = M^k$ anc $sqrt[k]c = N$. Then $x -c = (M^k - N^k)(M^{k-1}+ ... N^{k-1})= (sqrt[k] x - sqrt[k] c)(sqrt[k] x^{k-1} +.... + sqrt[k]c^{k-1})$
$endgroup$
– fleablood
Dec 5 '18 at 1:32




$begingroup$
"I don't think any similar rule applies for non-integers powers in terms of factorising polynomials" Sure there are. Just take $sqrt[k] x = M$ and $x = M^k$ anc $sqrt[k]c = N$. Then $x -c = (M^k - N^k)(M^{k-1}+ ... N^{k-1})= (sqrt[k] x - sqrt[k] c)(sqrt[k] x^{k-1} +.... + sqrt[k]c^{k-1})$
$endgroup$
– fleablood
Dec 5 '18 at 1:32










3 Answers
3






active

oldest

votes


















1












$begingroup$

Replace $x$ with $(x^{frac 13})^3$ and $c = (c^{frac 13})^3$ and so



$x - c = (x^{frac 13})^3 - (c^{frac 13})^3 = (x^{frac 13} - c^{frac 13})((x^{frac 13})^2 + x^{frac 13}y^{frac 13} + (c^{frac 13})^2)$



So



$(x -c) frac 1{x^{frac 23} + x^{frac 13}y^{frac 13} + c^{frac 23}}= x^{frac 13}- c^{frac 13}$



So $y = frac 1{x^{frac 23} + x^{frac 13}y^{frac 13} + c^{frac 23}}$



There IS the assumption that $x^{frac 23} + x^{frac 13}y^{frac 13} + c^{frac 23} ne 0$. But if $x^{frac 23} + x^{frac 13}y^{frac 13} + c^{frac 23} = 0$ then $x - c = (x^{frac 13})^3 - (c^{frac 13})^3 = (x^{frac 13} - c^{frac 13})((x^{frac 13})^2 + x^{frac 13}y^{frac 13} + (c^{frac 13})^2) = 0$ and $x = c$ and $x^{frac 13} = c^{frac 13}$ and $y$ could be anything.






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    Yes your way is right, let use that



    $$A^3-B^3=(A-B)(A^2+AB+B^2)$$



    with $A^3=x$ and $B^3=c$.



    Then



    $$x^{1/3}-c^{1/3}=(x-c)cdot frac1{x^{2/3}+x^{1/3}c^{1/3}+c^{2/3}}$$






    share|cite|improve this answer











    $endgroup$





















      -1












      $begingroup$

      Write $w=x-c$. You want to expand in powers of $w$. So, as $w to 0$ we have
      begin{align}
      x^{1/3} &= (c+w)^{1/3} = c^{1/3}left(1+frac{w}{c}right)^{1/3}
      \
      &= c^{1/3}left(1+frac{w}{3c}-frac{w^2}{9c^2}+frac{5w^3}{81c^3}+dotsright)
      \
      &= c^{1/3}+frac{w}{3c^{2/3}}-frac{w^2}{9c^{5/3}}+frac{5w^3}{81c^{8/3}}+dots
      \
      x^{1/3} - c^{1/3} &= frac{w}{3c^{2/3}}-frac{w^2}{9c^{5/3}}+frac{5w^3}{81c^{8/3}}+dots
      \
      &= wleft(frac{1}{3c^{2/3}}-frac{w}{9c^{5/3}}+frac{5w^2}{81c^{8/3}}+dotsright)
      \
      &= (x-c)left(frac{1}{3c^{2/3}}-frac{x-c}{9c^{5/3}}+frac{5(x-c)^2}{81c^{8/3}}+dotsright)
      end{align}



      Of course the last factor is exactly
      $$
      frac{x^{1/3}-c^{1/3}}{x-c} ,
      $$
      which is an analytic function (removable singularity) near $x=c$.






      share|cite|improve this answer











      $endgroup$









      • 1




        $begingroup$
        Good God! Really?
        $endgroup$
        – fleablood
        Dec 5 '18 at 1:34











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      3 Answers
      3






      active

      oldest

      votes








      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      1












      $begingroup$

      Replace $x$ with $(x^{frac 13})^3$ and $c = (c^{frac 13})^3$ and so



      $x - c = (x^{frac 13})^3 - (c^{frac 13})^3 = (x^{frac 13} - c^{frac 13})((x^{frac 13})^2 + x^{frac 13}y^{frac 13} + (c^{frac 13})^2)$



      So



      $(x -c) frac 1{x^{frac 23} + x^{frac 13}y^{frac 13} + c^{frac 23}}= x^{frac 13}- c^{frac 13}$



      So $y = frac 1{x^{frac 23} + x^{frac 13}y^{frac 13} + c^{frac 23}}$



      There IS the assumption that $x^{frac 23} + x^{frac 13}y^{frac 13} + c^{frac 23} ne 0$. But if $x^{frac 23} + x^{frac 13}y^{frac 13} + c^{frac 23} = 0$ then $x - c = (x^{frac 13})^3 - (c^{frac 13})^3 = (x^{frac 13} - c^{frac 13})((x^{frac 13})^2 + x^{frac 13}y^{frac 13} + (c^{frac 13})^2) = 0$ and $x = c$ and $x^{frac 13} = c^{frac 13}$ and $y$ could be anything.






      share|cite|improve this answer









      $endgroup$


















        1












        $begingroup$

        Replace $x$ with $(x^{frac 13})^3$ and $c = (c^{frac 13})^3$ and so



        $x - c = (x^{frac 13})^3 - (c^{frac 13})^3 = (x^{frac 13} - c^{frac 13})((x^{frac 13})^2 + x^{frac 13}y^{frac 13} + (c^{frac 13})^2)$



        So



        $(x -c) frac 1{x^{frac 23} + x^{frac 13}y^{frac 13} + c^{frac 23}}= x^{frac 13}- c^{frac 13}$



        So $y = frac 1{x^{frac 23} + x^{frac 13}y^{frac 13} + c^{frac 23}}$



        There IS the assumption that $x^{frac 23} + x^{frac 13}y^{frac 13} + c^{frac 23} ne 0$. But if $x^{frac 23} + x^{frac 13}y^{frac 13} + c^{frac 23} = 0$ then $x - c = (x^{frac 13})^3 - (c^{frac 13})^3 = (x^{frac 13} - c^{frac 13})((x^{frac 13})^2 + x^{frac 13}y^{frac 13} + (c^{frac 13})^2) = 0$ and $x = c$ and $x^{frac 13} = c^{frac 13}$ and $y$ could be anything.






        share|cite|improve this answer









        $endgroup$
















          1












          1








          1





          $begingroup$

          Replace $x$ with $(x^{frac 13})^3$ and $c = (c^{frac 13})^3$ and so



          $x - c = (x^{frac 13})^3 - (c^{frac 13})^3 = (x^{frac 13} - c^{frac 13})((x^{frac 13})^2 + x^{frac 13}y^{frac 13} + (c^{frac 13})^2)$



          So



          $(x -c) frac 1{x^{frac 23} + x^{frac 13}y^{frac 13} + c^{frac 23}}= x^{frac 13}- c^{frac 13}$



          So $y = frac 1{x^{frac 23} + x^{frac 13}y^{frac 13} + c^{frac 23}}$



          There IS the assumption that $x^{frac 23} + x^{frac 13}y^{frac 13} + c^{frac 23} ne 0$. But if $x^{frac 23} + x^{frac 13}y^{frac 13} + c^{frac 23} = 0$ then $x - c = (x^{frac 13})^3 - (c^{frac 13})^3 = (x^{frac 13} - c^{frac 13})((x^{frac 13})^2 + x^{frac 13}y^{frac 13} + (c^{frac 13})^2) = 0$ and $x = c$ and $x^{frac 13} = c^{frac 13}$ and $y$ could be anything.






          share|cite|improve this answer









          $endgroup$



          Replace $x$ with $(x^{frac 13})^3$ and $c = (c^{frac 13})^3$ and so



          $x - c = (x^{frac 13})^3 - (c^{frac 13})^3 = (x^{frac 13} - c^{frac 13})((x^{frac 13})^2 + x^{frac 13}y^{frac 13} + (c^{frac 13})^2)$



          So



          $(x -c) frac 1{x^{frac 23} + x^{frac 13}y^{frac 13} + c^{frac 23}}= x^{frac 13}- c^{frac 13}$



          So $y = frac 1{x^{frac 23} + x^{frac 13}y^{frac 13} + c^{frac 23}}$



          There IS the assumption that $x^{frac 23} + x^{frac 13}y^{frac 13} + c^{frac 23} ne 0$. But if $x^{frac 23} + x^{frac 13}y^{frac 13} + c^{frac 23} = 0$ then $x - c = (x^{frac 13})^3 - (c^{frac 13})^3 = (x^{frac 13} - c^{frac 13})((x^{frac 13})^2 + x^{frac 13}y^{frac 13} + (c^{frac 13})^2) = 0$ and $x = c$ and $x^{frac 13} = c^{frac 13}$ and $y$ could be anything.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 5 '18 at 1:24









          fleabloodfleablood

          69.3k22685




          69.3k22685























              0












              $begingroup$

              Yes your way is right, let use that



              $$A^3-B^3=(A-B)(A^2+AB+B^2)$$



              with $A^3=x$ and $B^3=c$.



              Then



              $$x^{1/3}-c^{1/3}=(x-c)cdot frac1{x^{2/3}+x^{1/3}c^{1/3}+c^{2/3}}$$






              share|cite|improve this answer











              $endgroup$


















                0












                $begingroup$

                Yes your way is right, let use that



                $$A^3-B^3=(A-B)(A^2+AB+B^2)$$



                with $A^3=x$ and $B^3=c$.



                Then



                $$x^{1/3}-c^{1/3}=(x-c)cdot frac1{x^{2/3}+x^{1/3}c^{1/3}+c^{2/3}}$$






                share|cite|improve this answer











                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  Yes your way is right, let use that



                  $$A^3-B^3=(A-B)(A^2+AB+B^2)$$



                  with $A^3=x$ and $B^3=c$.



                  Then



                  $$x^{1/3}-c^{1/3}=(x-c)cdot frac1{x^{2/3}+x^{1/3}c^{1/3}+c^{2/3}}$$






                  share|cite|improve this answer











                  $endgroup$



                  Yes your way is right, let use that



                  $$A^3-B^3=(A-B)(A^2+AB+B^2)$$



                  with $A^3=x$ and $B^3=c$.



                  Then



                  $$x^{1/3}-c^{1/3}=(x-c)cdot frac1{x^{2/3}+x^{1/3}c^{1/3}+c^{2/3}}$$







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Dec 5 '18 at 1:29

























                  answered Dec 5 '18 at 1:18









                  gimusigimusi

                  92.8k94494




                  92.8k94494























                      -1












                      $begingroup$

                      Write $w=x-c$. You want to expand in powers of $w$. So, as $w to 0$ we have
                      begin{align}
                      x^{1/3} &= (c+w)^{1/3} = c^{1/3}left(1+frac{w}{c}right)^{1/3}
                      \
                      &= c^{1/3}left(1+frac{w}{3c}-frac{w^2}{9c^2}+frac{5w^3}{81c^3}+dotsright)
                      \
                      &= c^{1/3}+frac{w}{3c^{2/3}}-frac{w^2}{9c^{5/3}}+frac{5w^3}{81c^{8/3}}+dots
                      \
                      x^{1/3} - c^{1/3} &= frac{w}{3c^{2/3}}-frac{w^2}{9c^{5/3}}+frac{5w^3}{81c^{8/3}}+dots
                      \
                      &= wleft(frac{1}{3c^{2/3}}-frac{w}{9c^{5/3}}+frac{5w^2}{81c^{8/3}}+dotsright)
                      \
                      &= (x-c)left(frac{1}{3c^{2/3}}-frac{x-c}{9c^{5/3}}+frac{5(x-c)^2}{81c^{8/3}}+dotsright)
                      end{align}



                      Of course the last factor is exactly
                      $$
                      frac{x^{1/3}-c^{1/3}}{x-c} ,
                      $$
                      which is an analytic function (removable singularity) near $x=c$.






                      share|cite|improve this answer











                      $endgroup$









                      • 1




                        $begingroup$
                        Good God! Really?
                        $endgroup$
                        – fleablood
                        Dec 5 '18 at 1:34
















                      -1












                      $begingroup$

                      Write $w=x-c$. You want to expand in powers of $w$. So, as $w to 0$ we have
                      begin{align}
                      x^{1/3} &= (c+w)^{1/3} = c^{1/3}left(1+frac{w}{c}right)^{1/3}
                      \
                      &= c^{1/3}left(1+frac{w}{3c}-frac{w^2}{9c^2}+frac{5w^3}{81c^3}+dotsright)
                      \
                      &= c^{1/3}+frac{w}{3c^{2/3}}-frac{w^2}{9c^{5/3}}+frac{5w^3}{81c^{8/3}}+dots
                      \
                      x^{1/3} - c^{1/3} &= frac{w}{3c^{2/3}}-frac{w^2}{9c^{5/3}}+frac{5w^3}{81c^{8/3}}+dots
                      \
                      &= wleft(frac{1}{3c^{2/3}}-frac{w}{9c^{5/3}}+frac{5w^2}{81c^{8/3}}+dotsright)
                      \
                      &= (x-c)left(frac{1}{3c^{2/3}}-frac{x-c}{9c^{5/3}}+frac{5(x-c)^2}{81c^{8/3}}+dotsright)
                      end{align}



                      Of course the last factor is exactly
                      $$
                      frac{x^{1/3}-c^{1/3}}{x-c} ,
                      $$
                      which is an analytic function (removable singularity) near $x=c$.






                      share|cite|improve this answer











                      $endgroup$









                      • 1




                        $begingroup$
                        Good God! Really?
                        $endgroup$
                        – fleablood
                        Dec 5 '18 at 1:34














                      -1












                      -1








                      -1





                      $begingroup$

                      Write $w=x-c$. You want to expand in powers of $w$. So, as $w to 0$ we have
                      begin{align}
                      x^{1/3} &= (c+w)^{1/3} = c^{1/3}left(1+frac{w}{c}right)^{1/3}
                      \
                      &= c^{1/3}left(1+frac{w}{3c}-frac{w^2}{9c^2}+frac{5w^3}{81c^3}+dotsright)
                      \
                      &= c^{1/3}+frac{w}{3c^{2/3}}-frac{w^2}{9c^{5/3}}+frac{5w^3}{81c^{8/3}}+dots
                      \
                      x^{1/3} - c^{1/3} &= frac{w}{3c^{2/3}}-frac{w^2}{9c^{5/3}}+frac{5w^3}{81c^{8/3}}+dots
                      \
                      &= wleft(frac{1}{3c^{2/3}}-frac{w}{9c^{5/3}}+frac{5w^2}{81c^{8/3}}+dotsright)
                      \
                      &= (x-c)left(frac{1}{3c^{2/3}}-frac{x-c}{9c^{5/3}}+frac{5(x-c)^2}{81c^{8/3}}+dotsright)
                      end{align}



                      Of course the last factor is exactly
                      $$
                      frac{x^{1/3}-c^{1/3}}{x-c} ,
                      $$
                      which is an analytic function (removable singularity) near $x=c$.






                      share|cite|improve this answer











                      $endgroup$



                      Write $w=x-c$. You want to expand in powers of $w$. So, as $w to 0$ we have
                      begin{align}
                      x^{1/3} &= (c+w)^{1/3} = c^{1/3}left(1+frac{w}{c}right)^{1/3}
                      \
                      &= c^{1/3}left(1+frac{w}{3c}-frac{w^2}{9c^2}+frac{5w^3}{81c^3}+dotsright)
                      \
                      &= c^{1/3}+frac{w}{3c^{2/3}}-frac{w^2}{9c^{5/3}}+frac{5w^3}{81c^{8/3}}+dots
                      \
                      x^{1/3} - c^{1/3} &= frac{w}{3c^{2/3}}-frac{w^2}{9c^{5/3}}+frac{5w^3}{81c^{8/3}}+dots
                      \
                      &= wleft(frac{1}{3c^{2/3}}-frac{w}{9c^{5/3}}+frac{5w^2}{81c^{8/3}}+dotsright)
                      \
                      &= (x-c)left(frac{1}{3c^{2/3}}-frac{x-c}{9c^{5/3}}+frac{5(x-c)^2}{81c^{8/3}}+dotsright)
                      end{align}



                      Of course the last factor is exactly
                      $$
                      frac{x^{1/3}-c^{1/3}}{x-c} ,
                      $$
                      which is an analytic function (removable singularity) near $x=c$.







                      share|cite|improve this answer














                      share|cite|improve this answer



                      share|cite|improve this answer








                      edited Dec 5 '18 at 1:38

























                      answered Dec 5 '18 at 1:28









                      GEdgarGEdgar

                      62.2k267168




                      62.2k267168








                      • 1




                        $begingroup$
                        Good God! Really?
                        $endgroup$
                        – fleablood
                        Dec 5 '18 at 1:34














                      • 1




                        $begingroup$
                        Good God! Really?
                        $endgroup$
                        – fleablood
                        Dec 5 '18 at 1:34








                      1




                      1




                      $begingroup$
                      Good God! Really?
                      $endgroup$
                      – fleablood
                      Dec 5 '18 at 1:34




                      $begingroup$
                      Good God! Really?
                      $endgroup$
                      – fleablood
                      Dec 5 '18 at 1:34


















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