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!
 A: "Hint": since $x+y=3$ then $3-x=y$ and hence $3x-x^2=xy=3.$ 
A: Hint: $$x^{ 4 }+y^{ 4 }={ \left( x^{ 2 }+y^{ 2 } \right)  }^{ 2 }-2{ x }^{ 2 }{ y }^{ 2 }={ \left( { \left( x+y \right)  }^{ 2 }-2xy \right)  }^{ 2 }-2{ x }^{ 2 }{ y }^{ 2 }$$
A: Let $a_n=x^n+y^n$.
Then
$$3a_n=(x+y)a_n=x^{n+1}+xy^n+yx^n+y^{n+1}=a_{n+1}+xya_{n-1}=a_{n+1}+3a_{n-1}$$
and so
$a_{n+1}=3a_n-3a_{n-1}$.
Also, $a_0=x^0+y^0=2$, $a_1=x+y=3$.
So $a_2=3a_1-3a_0=3$, $a_3=3a_2-3a_1=0$, and so on.
A: We have $$81=3^4=(x+y)^4\\=x^4+4x^3y+6x^2y^2+4xy^3+y^4\\=x^4+12x^2+54+12y^2+y^4\\=x^4+y^4+12(x^2+y^2)$$so once you solve for $x^2+y^2$, it should be straight-forward to solve for $x^4+y^4$.
A: By Vieta's formulas $x$ and $y$ are roots of the polynomial
$$ p(z)=z^2-3z+3 $$
and if $z\in\{x,y\}$ we have $z^2=3z-3$, from which $z^4=9z^2-18z+9=9z-18$ and
$$ x^4+y^4 = 9(x+y)-36 = \color{red}{-9}.$$

In general, the characteristic polynomial of the sequence $\{s_n=x^n+y^n\}_{n\geq 0}$ is exactly $p(z)$,
so $s_{n+2}=3(s_{n+1}-s_n)$. Since $s_0=2$ and $s_1=3$, it follows that $s_2=3(3-2)=3$ and
$s_3 = 3(3-3)=0$, so $s_4=3(0-3)=-9$ and $s_5=3(-9-0)=-27$, for instance.
A: $$(x+y)^2=3^2$$
$$x^2+2xy+y^2=9$$
$$x^2+y^2=9-2xy$$
square both sides
$$x^4+2x^2y^2+y^4=(9-2xy)^2$$
$$x^4+y^4=(9-2xy)^2-2x^2y^2=\color{red}{-9}$$
A: 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
$$
x^4+y^4=9e^{2i\pi/3}+9e^{-2i\pi/3}=18\cos\frac{2\pi}{3}=-\frac{1}{2}\cdot18=-9
$$
