finding $x^6+y^6$ given $x+y$ and $xy$ Here is the question
If $x + y = 4$ and $xy = 2$, then find $x^6+ y^6$.
This is from a previous timed competition, so fastest answers are the best answers.
I've tried using sum of cubes, but I dont know what to do after $(x^2+y^2)(x^4-x^2y^2+y^4)$
. The only other way I can think of is solving for x and y, but that wouldn't be too quick. Any help?
 A: Since $(x,y)$ are roots of $\ x^2-sx+p=0$
Then consider this equation as the characteristic equation of the linear induction relation $$U_{n+1}=sU_u-pU_{n-1}$$
Which solution is $\ U_n=x^n+y^n\ $ with initial values $\begin{cases}U_0=2\\U_1=4\end{cases}$
Then go on calculating successive values:

*

*$U_2=4U_1-2U_0=12$

*$U_3=4U_2-2U_1=40$

*$U_4=4U_3-2U_2=136$

*$U_5=4U_4-2U_3=464$

*$U_6=4U_5-2U_4=1584$
The advantage of this method, is that it is a no-brainer even if you are asked to get $x^{17}+y^{17}$ for instance, it is quite easy to calculate more terms.
A: $x^2+y^2=(x+y)^2-2xy=16-4=12.$
$x^4+y^4=(x^2+y^2)^2-2(xy)^2=12^2-2\cdot 2^2=144-8=136$.
Now you can use your factorization:
$x^6+y^6=(x^2+y^2)[(x^4+y^4)-(xy)^2]$.
A: Another way is to write out
$$ (x+y)^3=(x^3+y^3)+3xy(x+y), $$
$$ (x^3+y^3)^2=(x^6+y^6)+2(xy)^3. $$
A: Your sum is$$\begin{align}\sum_\pm(2\pm\sqrt{2})^6&=2(2^6+\sqrt{2}^6+15(2^4\sqrt{2}^2+2^2\sqrt{2}^4))\\&=2(64+8+15(2\sqrt{2})^2(2^2+2))\\&=2(72+15\cdot8\cdot6)\\&=1584.\end{align}$$
A: $(x+y) ^2 =16$$\Rightarrow$$ x^2 +y^2 +2xy=16$$\Rightarrow$$ x^2 +y^2 =12$
Because $xy=2$
$x^2 +y^2 =12$ $\Rightarrow $ $ (x^2 +y^2)^3 =12^3$$\Rightarrow $$x^6+y^6+3(xy)^2(x^2 +y^2) =1728$$\Rightarrow $$x^6+y^6=1728 - 3(4)(12)=1584$
Finally :
$x^6+y^6 =1584$
