1. ## factorization problem

If $[(x/\sqrt{y})^3+(y/\sqrt{x})^3]^2+[(x/\sqrt{y})^3-(y/\sqrt{x})^3]^2=10(x^3+y^3)..............(1)$

and $(\frac{x^6-y^6}{6x^4y^4})^3=x^3+y^3.........(2)$

then calculate $\frac{3x^3y^3}{y^3-x^3}$

Any suggestion of how to get to the solution will be appreciated.
Note: the final answer not necessary has to be an integer(could be a simpler form)

2. ## Re: factorization problem

Originally Posted by rochosh
If $[(x/\sqrt{y})^3+(y/\sqrt{x})^3]^2+[(x/\sqrt{y})^3-(y/\sqrt{x})^3]^2=10(x^3+y^3)..............(1)$

and $(\frac{x^6-y^6}{6x^4y^4})^3=x^3+y^3.........(2)$

then calculate $\frac{3x^3y^3}{y^3-x^3}$

Any suggestion of how to get to the solution will be appreciated.
Note: the final answer not necessary has to be an integer(could be a simpler form)

3. ## Re: factorization problem

Solving the first equation $(1)$ further enough you'll get:
$(x^3+y^3)^2=8x^3y^3...........(I)$

Factorizing and reducing the second one $(2)$:
$(x^3-y^3)^3(x^3+y^3)^2=(6x^4y^4)^3..........(II)$

Replacing $(I)$ in $(II)$:
$(x^3-y^3)^{3}8x^3y^3=(6x^4y^4)^3$

Solving this final equation in order to find the required one above and you'll get
$-1=\frac{3x^3y^3}{y^3-x^3}$

Reorganizing my attempt to post it, I casually found out the solution.
In any case another way to solve this problem will be appreciated.
Thanks for the remarking.

4. ## Re: factorization problem

Originally Posted by rochosh
Solving the first equation $(1)$ further enough you'll get:
$(x^3+y^3)^2=8x^3y^3...........(I)$

Factorizing and reducing the second one $(2)$:
$(x^3-y^3)^3(x^3+y^3)^2=(6x^4y^4)^3..........(II)$

Replacing $(I)$ in $(II)$:
$(x^3-y^3)^{3}8x^3y^3=(6x^4y^4)^3$

Solving this final equation in order to find the required one above and you'll get
$-1=\frac{3x^3y^3}{y^3-x^3}$

Reorganizing my attempt to post it, I casually found out the solution.
In any case another way to solve this problem will be appreciated.
Thanks for the remarking.
see i told you it works!