Simplyfying $(x^2-y^2)^frac12times(x-y)^frac32times(x+y)^frac-12$ algebraic experssion Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Problems and Resources to self-study medium level mathHow to simplify an equation of the $(x^2 + y^2)^1/2$ For example:Ascertaining a from logarithmic equationsCubic polynomial missing the linear termCoverting denary numbers into octal, binary and hexadecimal formStumped by a pretty basic fraction divisionHow do we get from $ln A=ln P+rn$ to $A=Pe^rn$ and similar logarithmic equations?How do I get from log F = log G + log m - log(1/M) - 2 log r to a solution withoug logs?Dividing higher-order algebraic expressionsSimplifying a polynomial

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Simplyfying $(x^2-y^2)^frac12times(x-y)^frac32times(x+y)^frac-12$ algebraic experssion



Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Problems and Resources to self-study medium level mathHow to simplify an equation of the $(x^2 + y^2)^1/2$ For example:Ascertaining a from logarithmic equationsCubic polynomial missing the linear termCoverting denary numbers into octal, binary and hexadecimal formStumped by a pretty basic fraction divisionHow do we get from $ln A=ln P+rn$ to $A=Pe^rn$ and similar logarithmic equations?How do I get from log F = log G + log m - log(1/M) - 2 log r to a solution withoug logs?Dividing higher-order algebraic expressionsSimplifying a polynomial










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I've been self-studying from Stroud & Booth's amazing "Engineering Mathematics", and am on the (precalculus) Algebra section. (And I've been asking lots of questions here as well!) I'm trying to simplify an algebraic expression, and I've tried it 101 ways, but I just can't seem to get a reasonable result. The polynomial is:



$$(x^2-y^2)^frac12times(x-y)^frac32times(x+y)^frac-12$$



Can anyone help me out? These are really making me feel dumb.










share|cite|improve this question









$endgroup$











  • $begingroup$
    Hint: $(x^2-y^2)=(x-y)cdot (x+y)$ I think you can finish.
    $endgroup$
    – callculus
    Mar 26 at 16:31
















0












$begingroup$


I've been self-studying from Stroud & Booth's amazing "Engineering Mathematics", and am on the (precalculus) Algebra section. (And I've been asking lots of questions here as well!) I'm trying to simplify an algebraic expression, and I've tried it 101 ways, but I just can't seem to get a reasonable result. The polynomial is:



$$(x^2-y^2)^frac12times(x-y)^frac32times(x+y)^frac-12$$



Can anyone help me out? These are really making me feel dumb.










share|cite|improve this question









$endgroup$











  • $begingroup$
    Hint: $(x^2-y^2)=(x-y)cdot (x+y)$ I think you can finish.
    $endgroup$
    – callculus
    Mar 26 at 16:31














0












0








0





$begingroup$


I've been self-studying from Stroud & Booth's amazing "Engineering Mathematics", and am on the (precalculus) Algebra section. (And I've been asking lots of questions here as well!) I'm trying to simplify an algebraic expression, and I've tried it 101 ways, but I just can't seem to get a reasonable result. The polynomial is:



$$(x^2-y^2)^frac12times(x-y)^frac32times(x+y)^frac-12$$



Can anyone help me out? These are really making me feel dumb.










share|cite|improve this question









$endgroup$




I've been self-studying from Stroud & Booth's amazing "Engineering Mathematics", and am on the (precalculus) Algebra section. (And I've been asking lots of questions here as well!) I'm trying to simplify an algebraic expression, and I've tried it 101 ways, but I just can't seem to get a reasonable result. The polynomial is:



$$(x^2-y^2)^frac12times(x-y)^frac32times(x+y)^frac-12$$



Can anyone help me out? These are really making me feel dumb.







algebra-precalculus self-learning






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Mar 26 at 16:29









neuronneuron

22917




22917











  • $begingroup$
    Hint: $(x^2-y^2)=(x-y)cdot (x+y)$ I think you can finish.
    $endgroup$
    – callculus
    Mar 26 at 16:31

















  • $begingroup$
    Hint: $(x^2-y^2)=(x-y)cdot (x+y)$ I think you can finish.
    $endgroup$
    – callculus
    Mar 26 at 16:31
















$begingroup$
Hint: $(x^2-y^2)=(x-y)cdot (x+y)$ I think you can finish.
$endgroup$
– callculus
Mar 26 at 16:31





$begingroup$
Hint: $(x^2-y^2)=(x-y)cdot (x+y)$ I think you can finish.
$endgroup$
– callculus
Mar 26 at 16:31











3 Answers
3






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$$(x^2-y^2)^frac12times(x-y)^frac32times(x+y)^frac-12$$
$$=(x-y)^frac12times(x+y)^frac12times(x-y)^frac32times(x+y)^frac-12$$
$$=(x-y)^frac12+frac32$$
$$=(x-y)^2.$$






share|cite|improve this answer









$endgroup$




















    1












    $begingroup$

    Since $x^2-y^2=(x+y)(x-y)$, $(x^2-y^2)^frac12=(x+y)^frac12(x-y)^frac12$. Therefore, your expression is equal to$$(x-y)^frac12+frac32=(x-y)^2.$$






    share|cite|improve this answer









    $endgroup$




















      0












      $begingroup$

      Write your term in the form
      $$fracsqrt(x-y)(x+y)sqrt(x-y)^3 sqrtx+y$$






      share|cite|improve this answer









      $endgroup$













        Your Answer








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        3 Answers
        3






        active

        oldest

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        3 Answers
        3






        active

        oldest

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        active

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        active

        oldest

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        0












        $begingroup$

        $$(x^2-y^2)^frac12times(x-y)^frac32times(x+y)^frac-12$$
        $$=(x-y)^frac12times(x+y)^frac12times(x-y)^frac32times(x+y)^frac-12$$
        $$=(x-y)^frac12+frac32$$
        $$=(x-y)^2.$$






        share|cite|improve this answer









        $endgroup$

















          0












          $begingroup$

          $$(x^2-y^2)^frac12times(x-y)^frac32times(x+y)^frac-12$$
          $$=(x-y)^frac12times(x+y)^frac12times(x-y)^frac32times(x+y)^frac-12$$
          $$=(x-y)^frac12+frac32$$
          $$=(x-y)^2.$$






          share|cite|improve this answer









          $endgroup$















            0












            0








            0





            $begingroup$

            $$(x^2-y^2)^frac12times(x-y)^frac32times(x+y)^frac-12$$
            $$=(x-y)^frac12times(x+y)^frac12times(x-y)^frac32times(x+y)^frac-12$$
            $$=(x-y)^frac12+frac32$$
            $$=(x-y)^2.$$






            share|cite|improve this answer









            $endgroup$



            $$(x^2-y^2)^frac12times(x-y)^frac32times(x+y)^frac-12$$
            $$=(x-y)^frac12times(x+y)^frac12times(x-y)^frac32times(x+y)^frac-12$$
            $$=(x-y)^frac12+frac32$$
            $$=(x-y)^2.$$







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Mar 26 at 16:48









            saket kumarsaket kumar

            167113




            167113





















                1












                $begingroup$

                Since $x^2-y^2=(x+y)(x-y)$, $(x^2-y^2)^frac12=(x+y)^frac12(x-y)^frac12$. Therefore, your expression is equal to$$(x-y)^frac12+frac32=(x-y)^2.$$






                share|cite|improve this answer









                $endgroup$

















                  1












                  $begingroup$

                  Since $x^2-y^2=(x+y)(x-y)$, $(x^2-y^2)^frac12=(x+y)^frac12(x-y)^frac12$. Therefore, your expression is equal to$$(x-y)^frac12+frac32=(x-y)^2.$$






                  share|cite|improve this answer









                  $endgroup$















                    1












                    1








                    1





                    $begingroup$

                    Since $x^2-y^2=(x+y)(x-y)$, $(x^2-y^2)^frac12=(x+y)^frac12(x-y)^frac12$. Therefore, your expression is equal to$$(x-y)^frac12+frac32=(x-y)^2.$$






                    share|cite|improve this answer









                    $endgroup$



                    Since $x^2-y^2=(x+y)(x-y)$, $(x^2-y^2)^frac12=(x+y)^frac12(x-y)^frac12$. Therefore, your expression is equal to$$(x-y)^frac12+frac32=(x-y)^2.$$







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Mar 26 at 16:31









                    José Carlos SantosJosé Carlos Santos

                    175k24134243




                    175k24134243





















                        0












                        $begingroup$

                        Write your term in the form
                        $$fracsqrt(x-y)(x+y)sqrt(x-y)^3 sqrtx+y$$






                        share|cite|improve this answer









                        $endgroup$

















                          0












                          $begingroup$

                          Write your term in the form
                          $$fracsqrt(x-y)(x+y)sqrt(x-y)^3 sqrtx+y$$






                          share|cite|improve this answer









                          $endgroup$















                            0












                            0








                            0





                            $begingroup$

                            Write your term in the form
                            $$fracsqrt(x-y)(x+y)sqrt(x-y)^3 sqrtx+y$$






                            share|cite|improve this answer









                            $endgroup$



                            Write your term in the form
                            $$fracsqrt(x-y)(x+y)sqrt(x-y)^3 sqrtx+y$$







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Mar 26 at 16:34









                            Dr. Sonnhard GraubnerDr. Sonnhard Graubner

                            79.1k42867




                            79.1k42867



























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