<s>[INST] Finding the value of $x-y$. Given two equations \begin{align} & x^4+y^4=\dfrac {-7}{9} \\ & x^3-y^3=3. \end{align} From these equations find the value of $(x-y).$
I have just factorize $x^3-y^3=(x-y)(x^2+xy+y^2)=3$. But don't know how to proceed from here. One way to find the values of $x$ and $y$ and then compute possible values of $x-y$. But that is a too long way. I think there is any simple way to find the value of $x-y$. Please help me to solve this.
 [/INST] Multiply equation (2) by $x-y$, giving $(x^4-x^3 y +xy^3 + y^4) = 3(x-y)$.
Next, (1) = -7/9, so
$-x^3 y +xy^3 - 7/9 = 3(x-y)$.
Factorise. 
$xy(-x^2 + y^2) - 7/9 = 3(x-y)$. 
$(y-x)(y+x)xy - 7/9 = 3(x-y)$.
$(x-y)(3+(y+x)xy) = - 7/9$
Using your idea, $x^2+xy+y^2 = 3/(x-y) = - 7/3*(3+(y+x)xy)$
Congrats, you now have a quadratic in $x$, or $y$, depending on how you feel.
Next we solve this to get $y$ in terms of $x$.
$3x^2 +3xy + 3y^2 = -21-7xy^2-7x^2y$
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