Tricky Finding

$\Rightarrow\dfrac{a}{a+b^2}+\dfrac{b}{b+a^2}=\dfrac78$

$\dfrac{a(b+a^2) + b(a+b^2)}{(a+b^2)(b+a^2)} = \dfrac{7}{8}$

$\dfrac{2ab+a^3+b^3}{ab+a^3+b^3+(ab)^2}=\dfrac{7}{8}$

Now, putting $ab=2$ .
$\dfrac{a^3+b^3+4}{a^3+b^3+6}=\dfrac{7}{8}$

Now, cross multiplying.
$8a^3+8b^3+32=7a^3+7b^3+42$
$a^3+b^3=10$
$\Rightarrow a^6+b^6=(a^3+b^3)^2-2(ab)^3$
$(10)^2-2×8$
$\boxed{84}$

Combine the fraction by creating a common denominator to get:

$\dfrac{a(b+a^2) + b(a+b^2)}{(a+b^2)(b+a^2)} = \dfrac{7}{8}$

Cross multiplying and simplifying (and using the fact that $ab = 2$ to remove all occurrences of $ab$ and $(ab)^2$ , we get $a^3 + b^3 = 10$ .

Squaring this, we get $a^6 + b^6 + 2a^3b^3 = 100$ . But we know that $ab = 2$ , so $2a^3b^3=16$ and the answer is:

$100 - 16 = \boxed{84}$