Evaluate $x+y=xy=3$ how would you evaluate $x^4+y^4$? I know how to evaluate $x^3 +y^3$ when $x+y=xy=3$, but how would you evaluate for? 

$$x^4+y^4$$

Any help would be appreciated.Thanks in advance!

You can express $x^4+y^4$ in terms of $s=x+y$ and $p=xy$:
$$
x^4+y^4=as^4+bs^2p+cp^2
$$
(the degree of $s$ is $1$ and the degree of $p$ is $2$). We can determine $a,b,c$ by using particular values for $x$ and $y$.


*

*If $x=1$ and $y=0$, then $x^4+y^4=1$, $s=1$ and $p=0$

*If $x=1$ and $y=1$, then $x^4+y^4=2$, $s=2$ and $p=1$

*If $x=1$ and $y=-1$, then $x^4+y^4=2$, $s=0$ and $p=-1$


\begin{cases}
1=a \\[4px]
2=16a+4b+c \\[4px]
2=c
\end{cases}
Thus $a=1$, $b=-4$ and $c=2$:
$$
x^4+y^4=(x+y)^4-4xy(x+y)^2+2(xy)^2
$$
(you can check the correctness by expanding the right-hand side).
In the case $x+y=xy=3$, we have
$$
x^4+y^4=3^4-4\cdot 3\cdot3^2+2\cdot3^2=-9
$$
A different approach is to compute $x$ and $y$, which are the roots of $z^2-3z+3=0$, so we can use
$$
x=\frac{3+i\sqrt{3}}{2},\qquad y=\bar{x}=\frac{3-i\sqrt{3}}{2}
$$
The modulus of $x$ and $y$ is $\sqrt{x\bar{x}}=\sqrt{3}$, so
$$
x=\sqrt{3}\left(\frac{\sqrt{3}}{2}+\frac{1}{2}i\right)
=\sqrt{3}e^{i\pi/6},
\qquad
y=\sqrt{3}\left(\frac{\sqrt{3}}{2}-\frac{1}{2}i\right)=
\sqrt{3}e^{-i\pi/6}
$$
Therefore
$$
x^4+y^4=9e^{2i\pi/3}+9e^{-2i\pi/3}=18\cos\frac{2\pi}{3}=-\frac{1}{2}\cdot18=-9
$$
