Solve the equation. \[ (x+1)^{2 / 5}-4(x+1)^{1 / 5}+3=0 \]

To solve the equation \[ (y)^{2} - 4(y) + 3 = 0 \] where we let \(y = (x + 1)^{1/5}\), we can rewrite the equation as \[ y^2 - 4y + 3 = 0. \] Now we can factor the quadratic: \[ (y - 1)(y - 3) = 0. \] Setting each factor equal to zero gives us: \[ y - 1 = 0 \quad \Rightarrow \quad y = 1, \] \[ y - 3 = 0 \quad \Rightarrow \quad y = 3. \] Now we convert back to our original variable \(x\): For \(y = 1\): \[ (x + 1)^{1/5} = 1 \quad \Rightarrow \quad x + 1 = 1^5 = 1 \quad \Rightarrow \quad x = 0. \] For \(y = 3\): \[ (x + 1)^{1/5} = 3 \quad \Rightarrow \quad x + 1 = 3^5 = 243 \quad \Rightarrow \quad x = 242. \] Thus, the solutions to the original equation are \[ \boxed{0} \quad \text{and} \quad \boxed{242}. \]