Question 1209473
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Evaluate{{{sqrt(y/x)}}}, where x and y are positive integers, and {{{0=x^5-x^3y^3 -12393}}}.
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D U P L I C A T E 



I solved this problem at this forum yesterday under this link


<A HREF=https://www.algebra.com/algebra/homework/expressions/expressions.faq.question.1209466.html>https://www.algebra.com/algebra/homework/expressions/expressions.faq.question.1209466.html</A>


https://www.algebra.com/algebra/homework/expressions/expressions.faq.question.1209466.html



For your convenience, I copy-paste this solution here again.



<pre>
            <U>S O L U T I O N</U>



Your starting equation is

    x^5 - x^3*y^3 = 12393,      (1)

or

    x^3*(x^2 - y^3) = 12393.    (2)


Integer number 12393 has the primary decomposition  12393 = {{{3^6*17}}},  
so the last equation is

    x^3*(x^2 - y^3) = {{{3^6*17}}}.   (3)


From it, it is clear that for x, y to be integer solutions to this equation, it is necessary that x be 1, or 3, or 3^2 = 9.


So, we should consider these three cases.


(a)  x = 1.  Then from equation (3)

     x^2 - y^3 = 3^6*17 = 12393,  y^3 = 1 - 12393 = -12392.

                                  But this number is not a positive perfect cube, so this way does not work.



(b)  x = 3.  Then from equation (3)

     x^2 - y^3 = 3^3*17 = 459,  y^3 = 9 - 459 = -450. 

                                  But this number is not a positive perfect cube, so this way does not work.



(c)  x = 3^2 = 9.  Then from equation (3)

     x^2 - y^3 = 17,  y^3 = 81 - 17 = 64. 

                                 This number, 64, is a positive perfect cube, so  y = 4.


Thus, the solution to the given equation in this pair of positive integer numbers  (x,y) = (3,4).


Then  {{{sqrt(y/x)}}}  is this irrational number {{{sqrt(4/3)}}} = {{{(2*sqrt(3))/3}}} = 1.154700538  (rounded).    <U>ANSWER</U>
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Solved.