Step 1 :Simplify the equation: \(\sqrt{t^3 + 1} = \left(\frac{y^4 + 1}{y^2 + 1}\right)^5\)
Step 2 :Solve for t in terms of y: \(t = \boxed{(-1 + \frac{(y^4 + 1)^{10}}{(y^2 + 1)^{10}})^{\frac{1}{3}}}\)