Let $a$ and $b$ be positive real numbers such that $ab=2$ and \[\dfrac{a}{a+b^2}+\dfrac{b}{b+a^2}=\dfrac78.\] Find $a^6+b^6$.

[i]Proposed by David Altizio[/i]
1. Given the equations:
   \[
   ab = 2
   \]
   and
   \[
   \dfrac{a}{a+b^2}+\dfrac{b}{b+a^2}=\dfrac{7}{8},
   \]
   we start by combining the fractions on the left-hand side.

2. Combine the fractions by creating a common denominator:
   \[
   \dfrac{a(b+a^2) + b(a+b^2)}{(a+b^2)(b+a^2)} = \dfrac{7}{8}.
   \]

3. Simplify the numerator:
   \[
   a(b + a^2) + b(a + b^2) = ab + a^3 + ba + b^3 = 2 + a^3 + b^3.
   \]
   Therefore, the equation becomes:
   \[
   \dfrac{2 + a^3 + b^3}{(a+b^2)(b+a^2)} = \dfrac{7}{8}.
   \]

4. Cross-multiply to clear the fraction:
   \[
   8(2 + a^3 + b^3) = 7(a+b^2)(b+a^2).
   \]

5. Expand the right-hand side:
   \[
   7(a+b^2)(b+a^2) = 7(ab + a^3 + b^3 + b^2a^2).
   \]
   Using \(ab = 2\), we get:
   \[
   7(2 + a^3 + b^3 + b^2a^2).
   \]

6. Substitute back into the equation:
   \[
   8(2 + a^3 + b^3) = 7(2 + a^3 + b^3 + b^2a^2).
   \]

7. Simplify and solve for \(a^3 + b^3\):
   \[
   16 + 8a^3 + 8b^3 = 14 + 7a^3 + 7b^3 + 14.
   \]
   \[
   16 + 8a^3 + 8b^3 = 14 + 7a^3 + 7b^3 + 14.
   \]
   \[
   16 + 8a^3 + 8b^3 = 28 + 7a^3 + 7b^3.
   \]
   \[
   8a^3 + 8b^3 - 7a^3 - 7b^3 = 28 - 16.
   \]
   \[
   a^3 + b^3 = 12.
   \]

8. Now, we need to find \(a^6 + b^6\). We start by squaring \(a^3 + b^3\):
   \[
   (a^3 + b^3)^2 = a^6 + b^6 + 2a^3b^3.
   \]
   Substituting \(a^3 + b^3 = 12\):
   \[
   12^2 = a^6 + b^6 + 2a^3b^3.
   \]
   \[
   144 = a^6 + b^6 + 2a^3b^3.
   \]

9. We know \(ab = 2\), so:
   \[
   a^3b^3 = (ab)^3 = 2^3 = 8.
   \]
   Therefore:
   \[
   2a^3b^3 = 2 \times 8 = 16.
   \]

10. Substitute \(2a^3b^3 = 16\) into the equation:
    \[
    144 = a^6 + b^6 + 16.
    \]
    \[
    a^6 + b^6 = 144 - 16.
    \]
    \[
    a^6 + b^6 = 128.
    \]

The final answer is \(\boxed{128}\).