Solve $\sqrt[4]{x}+\sqrt[4]{x+1}=\sqrt[4]{2x+1}$ Solve $\sqrt[4]{x}+\sqrt[4]{x+1}=\sqrt[4]{2x+1}$
My attempt:
Square both sides three times
$$\begin{align*}
36(x^2+x)&=4(\sqrt{x^2+x})(2x+1+\sqrt{x^2+x})\\
(\sqrt{x^2+x})(35\sqrt{x^2+x}-4(2x+1))&=0
\end{align*}$$
This means $0,-1$ are solutions but I can't make sure that these are the only solutions. Also I'm not sure that squaring three times is a good approach or not.

You have $x^{1/4}+(x+1)^{1/4}=(2x+1)^{1/4}$
Raise both sides to the power $4$ and you have:
$$x+(x+1)+4x^{3/4}(x+1)^{1/4}+6x^{1/4}(x+1)^{1/4}+4x^{1/4}(x+1)^{3/4}=2x+1$$
$$x^{1/4}(x+1)^{1/4}\Big(2x^{1/2}+3x^{1/4}(x+1)^{1/4}+2(x+1)^{1/4}\Big)=0$$
From here we have $x=0$ or $x=-1$ as solution.
Now if $x\neq 0$ and $x\neq -1$, then $2x^{1/2}+3x^{1/4}(x+1)^{1/4}+2(x+1)^{1/4}=0$
Observe that $2x^{1/2}+3x^{1/4}(x+1)^{1/4}+2(x+1)^{1/4}=2(x^{1/2}+2x^{1/4}(x+1)^{1/4}+(x+1)^{1/2})-x^{1/4}(x+1)^{1/4}=2\Big(x^{1/4}+(x+1)^{1/4}\Big)^2-x^{1/4}(x+1)^{1/4}$
$$2\Big(x^{1/4}+(x+1)^{1/4}\Big)^2=x^{1/4}(x+1)^{1/4}$$
$$2(2x+1)^{1/2}=x^{1/4}(x+1)^{1/4}$$
Raising both the sides to the power of $4$ gives:
$$16(2x+1)^2=x(x+1)$$
This in fact will give us an quadratic equation. I hope you can solve from here.
