$a+b=c+d$ , $a^3+b^3=c^3+d^3$ , prove that $a^{2009}+b^{2009}=c^{2009}+d^{2009}$












3












$begingroup$


I've got $a^2+b^2-ab=c^2+d^2-cd$.
I tried squaring or cubing it repeatedly but I didn't get what I wanted.
Now how do I proceed?






algebra-precalculus







share|cite|improve this question











$endgroup$












  • $begingroup$
    This is very probably a duplicate. Hint : try to express $a^n+b^n$ (for odd $n$) in terms of $a+b$ and $a^3+b^3$
    $endgroup$
    – Ewan Delanoy
    Jan 20 at 10:28






  • 1




    $begingroup$
    @Arthur $a^3+b^3=(a+b)(a^2+b^2-ab)$
    $endgroup$
    – user289143
    Jan 20 at 10:34






  • 2




    $begingroup$
    Welcome to MSE. You should choose your tags carefully. What has this to do with linear-algebra?
    $endgroup$
    – José Carlos Santos
    Jan 20 at 10:35
















3












$begingroup$


I've got $a^2+b^2-ab=c^2+d^2-cd$.
I tried squaring or cubing it repeatedly but I didn't get what I wanted.
Now how do I proceed?






algebra-precalculus







share|cite|improve this question











$endgroup$












  • $begingroup$
    This is very probably a duplicate. Hint : try to express $a^n+b^n$ (for odd $n$) in terms of $a+b$ and $a^3+b^3$
    $endgroup$
    – Ewan Delanoy
    Jan 20 at 10:28






  • 1




    $begingroup$
    @Arthur $a^3+b^3=(a+b)(a^2+b^2-ab)$
    $endgroup$
    – user289143
    Jan 20 at 10:34






  • 2




    $begingroup$
    Welcome to MSE. You should choose your tags carefully. What has this to do with linear-algebra?
    $endgroup$
    – José Carlos Santos
    Jan 20 at 10:35














3












3








3





$begingroup$


I've got $a^2+b^2-ab=c^2+d^2-cd$.
I tried squaring or cubing it repeatedly but I didn't get what I wanted.
Now how do I proceed?






algebra-precalculus







share|cite|improve this question











$endgroup$




I've got $a^2+b^2-ab=c^2+d^2-cd$.
I tried squaring or cubing it repeatedly but I didn't get what I wanted.
Now how do I proceed?






algebra-precalculus




algebra-precalculus






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 20 at 11:07









scrutari

1679




1679










asked Jan 20 at 10:25









RiyaRiya

211




211












  • $begingroup$
    This is very probably a duplicate. Hint : try to express $a^n+b^n$ (for odd $n$) in terms of $a+b$ and $a^3+b^3$
    $endgroup$
    – Ewan Delanoy
    Jan 20 at 10:28






  • 1




    $begingroup$
    @Arthur $a^3+b^3=(a+b)(a^2+b^2-ab)$
    $endgroup$
    – user289143
    Jan 20 at 10:34






  • 2




    $begingroup$
    Welcome to MSE. You should choose your tags carefully. What has this to do with linear-algebra?
    $endgroup$
    – José Carlos Santos
    Jan 20 at 10:35


















  • $begingroup$
    This is very probably a duplicate. Hint : try to express $a^n+b^n$ (for odd $n$) in terms of $a+b$ and $a^3+b^3$
    $endgroup$
    – Ewan Delanoy
    Jan 20 at 10:28






  • 1




    $begingroup$
    @Arthur $a^3+b^3=(a+b)(a^2+b^2-ab)$
    $endgroup$
    – user289143
    Jan 20 at 10:34






  • 2




    $begingroup$
    Welcome to MSE. You should choose your tags carefully. What has this to do with linear-algebra?
    $endgroup$
    – José Carlos Santos
    Jan 20 at 10:35
















$begingroup$
This is very probably a duplicate. Hint : try to express $a^n+b^n$ (for odd $n$) in terms of $a+b$ and $a^3+b^3$
$endgroup$
– Ewan Delanoy
Jan 20 at 10:28




$begingroup$
This is very probably a duplicate. Hint : try to express $a^n+b^n$ (for odd $n$) in terms of $a+b$ and $a^3+b^3$
$endgroup$
– Ewan Delanoy
Jan 20 at 10:28




1




1




$begingroup$
@Arthur $a^3+b^3=(a+b)(a^2+b^2-ab)$
$endgroup$
– user289143
Jan 20 at 10:34




$begingroup$
@Arthur $a^3+b^3=(a+b)(a^2+b^2-ab)$
$endgroup$
– user289143
Jan 20 at 10:34




2




2




$begingroup$
Welcome to MSE. You should choose your tags carefully. What has this to do with linear-algebra?
$endgroup$
– José Carlos Santos
Jan 20 at 10:35




$begingroup$
Welcome to MSE. You should choose your tags carefully. What has this to do with linear-algebra?
$endgroup$
– José Carlos Santos
Jan 20 at 10:35










3 Answers
3






active

oldest

votes


















2












$begingroup$

$$(a+b)^3=(c+d)^3$$



$$iff a^3+b^3+3ab(a+b)=c^3+d^3+3cd(c+d)$$



If $a+bne0, ab=cd (1)$



$iffdfrac ad=dfrac cb=k$(say) $ (2)$



$a+b=c+dimplies dk+b=bk+diff d(k-1)=b(k-1)$



Either $d=biff a=c$



or $k=1$ use this in $(2)$



Or Using $(1), ab=cda+b=c+d$ So if $a,b$ are roots of $t^2+bt+c=0,$



$c,d$ will be the roots of same equation $implies{a,b}equiv{c,d}$



In both bases we can prove $$a^n+b^n=c^n+d^n$$






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    Hint: the case $a+b=c+d=0$ is obvious, so you can assume $a+b=c+dne0$; prove that $ab=cd$. Further hint: $x^2-xy+y^2=(x+y)^2-{color{red}{?}}$.






    share|cite|improve this answer









    $endgroup$





















      0












      $begingroup$

      Use that we get from $$a+b=c+d$$ so $$a^3+b^3+3ab(a+b)=c^3+d^3+3cd(c+d)$$ and $$x^3+y^3=(x+y)(x^2-xy+y^2)$$






      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









        2












        $begingroup$

        $$(a+b)^3=(c+d)^3$$



        $$iff a^3+b^3+3ab(a+b)=c^3+d^3+3cd(c+d)$$



        If $a+bne0, ab=cd (1)$



        $iffdfrac ad=dfrac cb=k$(say) $ (2)$



        $a+b=c+dimplies dk+b=bk+diff d(k-1)=b(k-1)$



        Either $d=biff a=c$



        or $k=1$ use this in $(2)$



        Or Using $(1), ab=cda+b=c+d$ So if $a,b$ are roots of $t^2+bt+c=0,$



        $c,d$ will be the roots of same equation $implies{a,b}equiv{c,d}$



        In both bases we can prove $$a^n+b^n=c^n+d^n$$






        share|cite|improve this answer









        $endgroup$


















          2












          $begingroup$

          $$(a+b)^3=(c+d)^3$$



          $$iff a^3+b^3+3ab(a+b)=c^3+d^3+3cd(c+d)$$



          If $a+bne0, ab=cd (1)$



          $iffdfrac ad=dfrac cb=k$(say) $ (2)$



          $a+b=c+dimplies dk+b=bk+diff d(k-1)=b(k-1)$



          Either $d=biff a=c$



          or $k=1$ use this in $(2)$



          Or Using $(1), ab=cda+b=c+d$ So if $a,b$ are roots of $t^2+bt+c=0,$



          $c,d$ will be the roots of same equation $implies{a,b}equiv{c,d}$



          In both bases we can prove $$a^n+b^n=c^n+d^n$$






          share|cite|improve this answer









          $endgroup$
















            2












            2








            2





            $begingroup$

            $$(a+b)^3=(c+d)^3$$



            $$iff a^3+b^3+3ab(a+b)=c^3+d^3+3cd(c+d)$$



            If $a+bne0, ab=cd (1)$



            $iffdfrac ad=dfrac cb=k$(say) $ (2)$



            $a+b=c+dimplies dk+b=bk+diff d(k-1)=b(k-1)$



            Either $d=biff a=c$



            or $k=1$ use this in $(2)$



            Or Using $(1), ab=cda+b=c+d$ So if $a,b$ are roots of $t^2+bt+c=0,$



            $c,d$ will be the roots of same equation $implies{a,b}equiv{c,d}$



            In both bases we can prove $$a^n+b^n=c^n+d^n$$






            share|cite|improve this answer









            $endgroup$



            $$(a+b)^3=(c+d)^3$$



            $$iff a^3+b^3+3ab(a+b)=c^3+d^3+3cd(c+d)$$



            If $a+bne0, ab=cd (1)$



            $iffdfrac ad=dfrac cb=k$(say) $ (2)$



            $a+b=c+dimplies dk+b=bk+diff d(k-1)=b(k-1)$



            Either $d=biff a=c$



            or $k=1$ use this in $(2)$



            Or Using $(1), ab=cda+b=c+d$ So if $a,b$ are roots of $t^2+bt+c=0,$



            $c,d$ will be the roots of same equation $implies{a,b}equiv{c,d}$



            In both bases we can prove $$a^n+b^n=c^n+d^n$$







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Jan 20 at 10:45









            lab bhattacharjeelab bhattacharjee

            226k15157275




            226k15157275























                0












                $begingroup$

                Hint: the case $a+b=c+d=0$ is obvious, so you can assume $a+b=c+dne0$; prove that $ab=cd$. Further hint: $x^2-xy+y^2=(x+y)^2-{color{red}{?}}$.






                share|cite|improve this answer









                $endgroup$


















                  0












                  $begingroup$

                  Hint: the case $a+b=c+d=0$ is obvious, so you can assume $a+b=c+dne0$; prove that $ab=cd$. Further hint: $x^2-xy+y^2=(x+y)^2-{color{red}{?}}$.






                  share|cite|improve this answer









                  $endgroup$
















                    0












                    0








                    0





                    $begingroup$

                    Hint: the case $a+b=c+d=0$ is obvious, so you can assume $a+b=c+dne0$; prove that $ab=cd$. Further hint: $x^2-xy+y^2=(x+y)^2-{color{red}{?}}$.






                    share|cite|improve this answer









                    $endgroup$



                    Hint: the case $a+b=c+d=0$ is obvious, so you can assume $a+b=c+dne0$; prove that $ab=cd$. Further hint: $x^2-xy+y^2=(x+y)^2-{color{red}{?}}$.







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Jan 20 at 10:30









                    egregegreg

                    183k1486205




                    183k1486205























                        0












                        $begingroup$

                        Use that we get from $$a+b=c+d$$ so $$a^3+b^3+3ab(a+b)=c^3+d^3+3cd(c+d)$$ and $$x^3+y^3=(x+y)(x^2-xy+y^2)$$






                        share|cite|improve this answer









                        $endgroup$


















                          0












                          $begingroup$

                          Use that we get from $$a+b=c+d$$ so $$a^3+b^3+3ab(a+b)=c^3+d^3+3cd(c+d)$$ and $$x^3+y^3=(x+y)(x^2-xy+y^2)$$






                          share|cite|improve this answer









                          $endgroup$
















                            0












                            0








                            0





                            $begingroup$

                            Use that we get from $$a+b=c+d$$ so $$a^3+b^3+3ab(a+b)=c^3+d^3+3cd(c+d)$$ and $$x^3+y^3=(x+y)(x^2-xy+y^2)$$






                            share|cite|improve this answer









                            $endgroup$



                            Use that we get from $$a+b=c+d$$ so $$a^3+b^3+3ab(a+b)=c^3+d^3+3cd(c+d)$$ and $$x^3+y^3=(x+y)(x^2-xy+y^2)$$







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Jan 20 at 10:34









                            Dr. Sonnhard GraubnerDr. Sonnhard Graubner

                            76.8k42866




                            76.8k42866






























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