Let $x,y$ be complex numbers such that $\dfrac{x^2+y^2}{x+y}=4$ and $\dfrac{x^4+y^4}{x^3+y^3}=2$.  Find all possible values of $\dfrac{x^6+y^6}{x^5+y^5}$.
1. Let \( a = \frac{x^2 + y^2}{xy} \) and \( b = \frac{x + y}{xy} \). From the first given equation, we have:
   \[
   \frac{x^2 + y^2}{x + y} = 4
   \]
   Multiplying both sides by \( \frac{1}{xy} \), we get:
   \[
   \frac{x^2 + y^2}{xy} \cdot \frac{1}{\frac{x + y}{xy}} = 4 \cdot \frac{1}{\frac{x + y}{xy}}
   \]
   Simplifying, we find:
   \[
   a = 4b
   \]

2. From the second given equation, we have:
   \[
   \frac{x^4 + y^4}{x^3 + y^3} = 2
   \]
   We can express \( x^4 + y^4 \) and \( x^3 + y^3 \) in terms of \( a \) and \( b \):
   \[
   x^4 + y^4 = (x^2 + y^2)^2 - 2x^2y^2 = a^2(xy)^2 - 2(xy)^2 = (a^2 - 2)(xy)^2
   \]
   \[
   x^3 + y^3 = (x + y)(x^2 - xy + y^2) = b(xy)(a - 1)
   \]
   Substituting these into the given equation, we get:
   \[
   \frac{(a^2 - 2)(xy)^2}{b(xy)(a - 1)} = 2
   \]
   Simplifying, we find:
   \[
   \frac{a^2 - 2}{b(a - 1)} = 2
   \]

3. Substituting \( a = 4b \) into the equation, we get:
   \[
   \frac{(4b)^2 - 2}{b(4b - 1)} = 2
   \]
   Simplifying, we find:
   \[
   \frac{16b^2 - 2}{4b^2 - b} = 2
   \]
   Multiplying both sides by \( 4b^2 - b \), we get:
   \[
   16b^2 - 2 = 2(4b^2 - b)
   \]
   Simplifying, we find:
   \[
   16b^2 - 2 = 8b^2 - 2b
   \]
   \[
   8b^2 = 2b
   \]
   \[
   4b^2 = b
   \]
   \[
   4b^2 - b = 0
   \]
   \[
   b(4b - 1) = 0
   \]
   Thus, \( b = 0 \) or \( b = \frac{1}{4} \).

4. Since \( b = 0 \) would make \( x + y = 0 \), which is not possible, we have:
   \[
   b = \frac{1}{4}
   \]

5. Now, we need to find \( \frac{x^6 + y^6}{x^5 + y^5} \). We can express \( x^6 + y^6 \) and \( x^5 + y^5 \) in terms of \( a \) and \( b \):
   \[
   x^6 + y^6 = (x^2 + y^2)(x^4 + y^4) - x^2y^2(x^2 + y^2) = a(xy)(a^2 - 2)(xy)^2 - (xy)^3a
   \]
   \[
   x^5 + y^5 = (x + y)(x^4 - xy(x^2 + y^2) + y^4) = b(xy)(a^2 - 2 - a)
   \]
   Substituting \( a = 4b \) and \( b = \frac{1}{4} \), we get:
   \[
   a = 4 \cdot \frac{1}{4} = 1
   \]
   \[
   x^6 + y^6 = (1)(xy)(1^2 - 2)(xy)^2 - (xy)^3(1) = -xy^3
   \]
   \[
   x^5 + y^5 = \frac{1}{4}(xy)(1^2 - 2 - 1) = -\frac{1}{4}xy
   \]
   Thus, we have:
   \[
   \frac{x^6 + y^6}{x^5 + y^5} = \frac{-xy^3}{-\frac{1}{4}xy} = 4
   \]

The final answer is \(\boxed{4}\).