Adding rational expr: 5 / (x^3 - y^3) + 3 / (x^2 + xy + y^2)

[alee]

The problem: 5 / (x^3 - y^3) + 3 / (x^2 + xy + y^2) = ?

My work so far: 5 / (x - y(x^2 - y^2)) + 3 / [(x + y)(x + y)]

I don't know what to do from here for a common denominator

thank you
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Edited by stapel -- Reason for edit: attempting to restore formatting and thus meaning

 

[galactus]

Is this your problem?. Try using LaTex. Your post got all mixed up. Hard to read.

\(\displaystyle \L\\\frac{5}{x^{3}-y^{3}}+\frac{3}{x^{2}+xy+y^{2}}\)

If this is correct, you should recognize the difference of two cubes.

\(\displaystyle \L\\x^{3}-y^{3}=(x-y)(x^{2}+xy+y^{2})\)

 

[Deleted member 4993]

Re: Adding rational expressions

5                  +                    3                             
-------------               ------------------   =                                          
x^3 - y^3      +         x^2 + xy + y^2

Are these two separate problems?

    5                    +              3
---------------          ---------------   = 
 x-y(x^2-y^2)    +     (x+y)(x+y)

 

[skeeter]

Re: Adding rational expressions

5 + 3
------------- ------------------ =
x^3 - y^3 + x^2 + xy + y^2



5 + 3
--------------- --------------- =
x-y(x^2-y^2) + (x+y)(x+y)


I don't know what to do from here for a common denominator

thank you

first, note that \(\displaystyle \L x^2 + xy + y^2 \neq (x+y)(x+y)\).
second, note galactus' factorization of \(\displaystyle \L x^3 - y^3\). the common denominator will be \(\displaystyle \L x^3 - y^3 = (x-y)(x^2 + xy + y^2)\) ...

\(\displaystyle \L \frac{5}{(x-y)(x^2 + xy + y^2)} + \frac{3(x-y)}{(x-y)(x^2 + xy + y^2)}\) ... add'em up.