I have the following equation:
$$\frac{\sqrt{n}+\sqrt{n+16(\sqrt{n}+1)}}{8} - \sqrt[3]{\frac{n}{4}} = 0\ \ \ \ \ \ \ \ \ (1).$$
In order to solve $(1)$, I am doing the following: $$(1) \implies \sqrt{n}+\sqrt{n+16(\sqrt{n}+1)} = 4 \sqrt[3]{2} \sqrt[3]{n}\ \ \ \ \ \ \ \ \ (2).$$ On squaring both sides of $(2)$ and rearranging, we get $$\sqrt{n} \sqrt{n+16(\sqrt{n}+1)} = 2^3 \sqrt[3]{4}\sqrt[3]{n^2} - (n + 8\sqrt{n}+8)\ \ \ \ \ \ \ \ \ (3).$$ On squaring both sides of $(3)$ and rearranging, we get $$\sqrt[3]{4}(n^{1/6})^{10}- 2^3 \sqrt[3]{2} (n^{1/6})^{8} + 2^3 \sqrt[3]{4} (n^{1/6})^{7} - 2^2 (n^{1/6})^6 + 2^3 \sqrt[3]{4}(n^{1/6})^{4} - 2^3 (n^{1/6})^{3} - 2^2 = 0\ \ \ \ \ \ \ \ \ \ \ (4).$$ On substituting $n^{1/6} = t$ in $(4)$, we get
$$\sqrt[3]{4}t^{10}- 2^3 \sqrt[3]{2} t^{8} + 2^3 \sqrt[3]{4} t^{7} - 2^2 t^6 + 2^3 \sqrt[3]{4}t^{4} - 2^3 t^{3} - 2^2 = 0\ \ \ \ \ \ \ \ \ \ \ (5).$$
Now, in order to solve $(5)$, I used Matlab. And I got the following roots of $(5)$:
$$-3.0256 + 0.0000i\\ 1.5465 + 0.5302i\\ 1.5465 - 0.5302i\\ 0.9641 + 0.0000i\\ -0.8112 + 0.0000i\\ -0.7359 + 0.0000i\\ 0.0814 + 0.8947i\\ 0.0814 - 0.8947i\\ 0.1764 + 0.7998i\\ 0.1764 - 0.7998i$$
Now, as we can see $t = 0.9641$ is a root of $(5)$, which implies, for $n^{1/6}= 0.9641$, equation $(4)$ is satisfied. Equivalently, we can say, for $n = 0.9641^{6} = 0.803031244 $, equation $(4)$ is satisfied. Therefore, for $n = 0.803031244$, equation $(1)$ should also satisfy. However, on substituting $n = 0.803031244$ in equation $(1)$, I am getting $$\frac{\sqrt{0.803031244}+\sqrt{0.803031244+16(\sqrt{0.803031244}+1)}}{8} - \sqrt[3]{\frac{0.803031244}{4}} = 0,$$ or equivalently, $$0.224024772 = 0\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (6).$$
Clearly, in the above equation, we can see that $L.H.S \neq R.H.S$. Can anybody shed some light? Why am I getting this huge difference between $L.H.S$ and $R.H.S$? Why $n = 0.803031244$ is not satisfying equation $(1)$?