Solve it $\cdots$

$\begin{aligned} \sqrt[3]{x+1} & = \sqrt{x-3} & \small \color{#3D99F6} \text{Let }y = x-3 \\ \sqrt[3]{y+4} & = \sqrt{y} & \small \color{#3D99F6} \text{Raise both sides by a power of }6 \\ (y+4)^2 & = y^3 \\ y^2 + 8y+16 & = y^3 \\ y^3 - y^2 - 8y-16 & =0 \\ (y-4)(y^2+3y+4) & =0 \\ y & = 4 & \small \color{#3D99F6} \text{Note that } y^2+3y+4 = 0 \text{ has no real root.} \\ \implies x & = \boxed{7} \end{aligned}$

${\sqrt[3]{x+1}}^6 ={\sqrt{x-3}}^6$ ${(x+1)}^2 = {(x-3)}^2$ $x^2+2x+1 = x^3-9x^2+27x-27$ $x^3-10x^2+25x- 8= 0$ ${\left(x-7\right)\left(x^{2\ }-3x+4\right)\ } = 0$ $\left(x^{2\ }-3x+4\right)>0$ so x=7