symmetric polynomial recursion to solve the system, $x^5+y^5=33$, $x+y=3$ I was just reading on symmetric polynomials and was given the system of equations$$x^5+y^5=33 \text{ , } x+y=3$$
In the text they said to denote $\sigma_1=x+y$ and $\sigma_2=xy$, and to use recursion.  I understand how the formula below works
$$S_5=x^5+y^5=\sigma_1 S_4-\sigma_2 S_3$$
But doing the recursion takes a long time.  Is there a shorter way to do the recursion to solve this specific problem?
Thanks
 A: Let $x+y=3=a$, $xy=b$, $x^5+y^5=33=c$.
\begin{align} 
x^5+y^5&=(x+y)(x^4-x^3y+x^2y^2-xy^3+y^4)
\\
c&=
a(x^4+y^4-x^2xy+(xy)^2-xyy^2)
\\
c&=
x^4+y^4-b(x^2+y^2)+b^2
,\\
x^4+y^4&=(x^2+y^2)^2-2x^2y^2=(x^2+y^2)^2-2b^2
,\\
c&=
a((x^2+y^2)^2-2b^2-b(x^2+y^2)+b^2)
=
a(x^2+y^2)(x^2+y^2-b)-b^2)
,\\
x^2+y^2&=(x+y)^2-2xy=a^2-2b
,\\
c&=
a(a^2-2b)(a^2-2b-b)-ab^2
,
\end{align}  
thus we have a quadratic equation in terms of $b$:
\begin{align} 
5ab^2-5a^3b-c+a^5&=0
,
\end{align}
\begin{align} 
b&=\frac{(5a^3\pm\sqrt{5a^6+20ac}}{10a}
,\\
b_1&=2
,\\
b_2&=7
.
\end{align}  
So, we need to consider two cases: $xy=2$ and $xy=7$.
In both cases we also have $x+y=3$.
Case 1 $xy=2$.
The quadratic equation with roots $x,y$ is
\begin{align} 
t^2-3t+2&=0
,\\
x,y&=1,2
.
\end{align}
Indeed, $1^5+2^5=33$, $1+2=3$, 
so in Case 1 we have two real solutions, $x=1,y=2$ and $x=2,y=1$. 
Case 2 $xy=7$.
The quadratic equation with roots $x,y$ is
\begin{align} 
t^2-3t+7&=0
,\\
x,y&=\tfrac32\pm\tfrac{\sqrt{19}}2\cdot i
,
\end{align}
so in Case 2 we have 
two complex conjugate solutions, 
\begin{align} 
x&=\tfrac32+\tfrac{\sqrt{19}}2\cdot i
,\\
y&=\tfrac32-\tfrac{\sqrt{19}}2\cdot i
\end{align}
and 
\begin{align} 
x&=\tfrac32-\tfrac{\sqrt{19}}2\cdot i
,\\
y&=\tfrac32+\tfrac{\sqrt{19}}2\cdot i 
.
\end{align} 
A: Also, we can use the homogenization.
Let $y=tx$.
Thus, $$t^5+1=\frac{33(t+1)^5}{243}$$ or
$$(t+1)(2t-1)(t-2)(7t^2+5t+7)=0$$ and the rest is smooth.
A: 
Is there a shorter way to do the recursion to solve this specific problem?`

If you don't insist on recursion, the shorter way in this case  is arguably the most direct one - just solve the system by eliminating one of the variables, then factoring with the rational root theorem:
$$
\begin{align}
0 = x^5 + y^5 - 33 &= x^5+(3-x)^5-33 \\
 &= 15 (x^4 - 6 x^3 + 18 x^2 - 27 x + 14) \\
 &= 15 (x - 1) (x - 2) (x^2 - 3 x + 7)
\end{align}
$$
