Solve the equation $\sqrt{3\sqrt[3]{x}+1}=\sqrt[3]{2\sqrt{x+1}-1}$

Question

Solve the equation $\sqrt{3\sqrt[3]{x}+1}=\sqrt[3]{2\sqrt{x+1}-1}$.

My attempt: With $u=\sqrt[3]{x}, v=\sqrt{x+1}$ I have $u^3=v^2-1$ and $(3u+1)^3=(2v-1)^2$ And I finally have a quadratic equation of u in terms of v. But the $\Delta$ is not $a^2$.

Any help is appreciated.

Answer 1

Let $\sqrt[3]x=a$.

Thus, we need to solve $$(3a+1)^3=\left(2\sqrt{a^3+1}-1\right)^2$$ or $$23a^3+27a^2+9a=4(1-\sqrt{a^3+1})$$ or $$a\left(23a^2+27a+9+\frac{4a^2}{1+\sqrt{a^3+1}}\right)=0,$$ which since $$27^2-4\cdot23\cdot9<0,$$ gives $$a=0$$ and $$x=0.$$