Let $a,b$ be real numbers such that $|a| \neq |b|$ and $\frac{a+b}{a-b}+\frac{a-b}{a+b}=6.$ Find the value of the expression $\frac{a^3+b^3}{a^3-b^3}+\frac{a^3-b^3}{a^3+b^3}.$
1. Given the equation:
   \[
   \frac{a+b}{a-b} + \frac{a-b}{a+b} = 6
   \]
   We start by simplifying the left-hand side. Using the identity for the sum of fractions:
   \[
   \frac{a+b}{a-b} + \frac{a-b}{a+b} = \frac{(a+b)^2 + (a-b)^2}{(a-b)(a+b)}
   \]
2. Expanding the squares in the numerator:
   \[
   (a+b)^2 = a^2 + 2ab + b^2
   \]
   \[
   (a-b)^2 = a^2 - 2ab + b^2
   \]
   Therefore:
   \[
   (a+b)^2 + (a-b)^2 = a^2 + 2ab + b^2 + a^2 - 2ab + b^2 = 2a^2 + 2b^2
   \]
3. The denominator is:
   \[
   (a-b)(a+b) = a^2 - b^2
   \]
   Thus, the equation becomes:
   \[
   \frac{2a^2 + 2b^2}{a^2 - b^2} = 6
   \]
4. Simplifying the fraction:
   \[
   \frac{2(a^2 + b^2)}{a^2 - b^2} = 6
   \]
   Dividing both sides by 2:
   \[
   \frac{a^2 + b^2}{a^2 - b^2} = 3
   \]
5. Cross-multiplying to solve for \(a^2\) and \(b^2\):
   \[
   a^2 + b^2 = 3(a^2 - b^2)
   \]
   \[
   a^2 + b^2 = 3a^2 - 3b^2
   \]
   \[
   4b^2 = 2a^2
   \]
   \[
   a^2 = 2b^2
   \]
   \[
   a = \sqrt{2}b \quad \text{or} \quad a = -\sqrt{2}b
   \]
6. Now, we need to find the value of:
   \[
   \frac{a^3 + b^3}{a^3 - b^3} + \frac{a^3 - b^3}{a^3 + b^3}
   \]
   Using the identity for the sum and difference of cubes:
   \[
   a^3 + b^3 = (a+b)(a^2 - ab + b^2)
   \]
   \[
   a^3 - b^3 = (a-b)(a^2 + ab + b^2)
   \]
7. Substituting \(a = \sqrt{2}b\):
   \[
   a^3 = (\sqrt{2}b)^3 = 2\sqrt{2}b^3
   \]
   \[
   b^3 = b^3
   \]
   Therefore:
   \[
   \frac{a^3 + b^3}{a^3 - b^3} = \frac{2\sqrt{2}b^3 + b^3}{2\sqrt{2}b^3 - b^3} = \frac{(2\sqrt{2} + 1)b^3}{(2\sqrt{2} - 1)b^3} = \frac{2\sqrt{2} + 1}{2\sqrt{2} - 1}
   \]
   Similarly:
   \[
   \frac{a^3 - b^3}{a^3 + b^3} = \frac{2\sqrt{2}b^3 - b^3}{2\sqrt{2}b^3 + b^3} = \frac{2\sqrt{2} - 1}{2\sqrt{2} + 1}
   \]
8. Adding these two fractions:
   \[
   \frac{2\sqrt{2} + 1}{2\sqrt{2} - 1} + \frac{2\sqrt{2} - 1}{2\sqrt{2} + 1}
   \]
   Using the identity for the sum of fractions:
   \[
   \frac{(2\sqrt{2} + 1)^2 + (2\sqrt{2} - 1)^2}{(2\sqrt{2} - 1)(2\sqrt{2} + 1)}
   \]
9. Simplifying the numerator:
   \[
   (2\sqrt{2} + 1)^2 = 8 + 4\sqrt{2} + 1 = 9 + 4\sqrt{2}
   \]
   \[
   (2\sqrt{2} - 1)^2 = 8 - 4\sqrt{2} + 1 = 9 - 4\sqrt{2}
   \]
   Therefore:
   \[
   (2\sqrt{2} + 1)^2 + (2\sqrt{2} - 1)^2 = 9 + 4\sqrt{2} + 9 - 4\sqrt{2} = 18
   \]
10. Simplifying the denominator:
    \[
    (2\sqrt{2} - 1)(2\sqrt{2} + 1) = 8 - 1 = 7
    \]
    Thus:
    \[
    \frac{18}{7} = \frac{18}{7}
    \]

The final answer is \(\boxed{\frac{18}{7}}\).