SOLUTION: 1. If X^2+x+1=0 ,then what is the value of (x^3+1/x^3)^3 ? 2. If X^X+Y=Y^3 and Y^X+Y=X^6*Y^3 ,where X and Y are natural numbers then what is the value of Y^X ?

Algebra ->  Equations -> SOLUTION: 1. If X^2+x+1=0 ,then what is the value of (x^3+1/x^3)^3 ? 2. If X^X+Y=Y^3 and Y^X+Y=X^6*Y^3 ,where X and Y are natural numbers then what is the value of Y^X ?       Log On


   



Question 596132: 1. If X^2+x+1=0 ,then what is the value of (x^3+1/x^3)^3 ?
2. If X^X+Y=Y^3 and Y^X+Y=X^6*Y^3 ,where X and Y are natural numbers then what is the value of Y^X ?


Answer by Edwin McCravy(20055) About Me  (Show Source):
You can put this solution on YOUR website!
I already answered 1

http://www.algebra.com/algebra/homework/equations/Equations.faq.question.596131.html 

2. If X^X+Y=Y^3 and Y^X+Y=X^6*Y^3 ,where X and Y are natural numbers then what
is the value of Y^X ?

There are no natural number solutions to that, so I assumed you must
have left off the necessary parentheses to show where the exponents start and
end.  So I rewrote it as this, which does have 2 natural number solutions:  

2. If X^(X+Y)=Y^3 and Y^(X+Y)=X^6*Y^3 ,where X and Y are natural numbers then
what is the value of Y^X ?

or in superscripted exponent notation:

2. If X%5E%28X%2BY%29 = Y%5E3 and Y%5E%28X%2BY%29 = X%5E6%2AY%5E3 ,where X and
Y are natural numbers then what is the value of Y^X ?

There are two solutions to that  (x,y) = (1,1) and (x,y) = (2,4)

So for the first solution y%5Ex%29 = 1%5E1 = 1
And for the second y%5Ex = 4%5E2 = 16

Edwin