nth Root Problems: Very Difficult Problems with Solutions

Solution:
[tex]\left(1+\sqrt{3}\right)^3=1+3\sqrt{3}+9+3\sqrt{3}=10+6\sqrt{3}[/tex], so [tex]\sqrt[3]{10+6\sqrt{3}}=1+\sqrt{3}[/tex].
[tex]3+\sqrt{3}+1+\sqrt{3}=4+2\sqrt{3}=\left(1+\sqrt{3}\right)^2[/tex]
[tex]\sqrt{\left(1+\sqrt{3}\right)^2}-\sqrt{3}=1+\sqrt{3}-\sqrt{3}=1[/tex]

Solution:
Let [tex]x=\sqrt[3]{2+\sqrt{5}}+\sqrt[3]{2-\sqrt{5}}[/tex]. Cube both sides using [tex](a+b)^3=a^3+b^3+3ab(a+b)[/tex]:
[tex]x^3=\left(2+\sqrt{5}\right)+\left(2-\sqrt{5}\right)+3\sqrt[3]{\left(2+\sqrt{5}\right)\left(2-\sqrt{5}\right)}\times x[/tex]
[tex]\sqrt[3]{4-5}=\sqrt[3]{-1}=-1[/tex], so [tex]x^3=4-3x[/tex].
[tex]x^3+3x-4=0 \Leftrightarrow (x-1)\left(x^2+x+4\right)=0[/tex]
The quadratic factor has a negative discriminant, so the only real root is [tex]x=1[/tex].
Another way: notice that [tex]\left(\frac{1+\sqrt{5}}{2}\right)^3=2+\sqrt{5}[/tex] and [tex]\left(\frac{1-\sqrt{5}}{2}\right)^3=2-\sqrt{5}[/tex].

Solution:
In the first bracket let [tex]u=\sqrt[3]{a}[/tex] ([tex]a=u^3[/tex]):
[tex]\frac{u^3-u}{u-1}-\frac{u^3+u^2}{u+1}=\frac{u(u-1)(u+1)}{u-1}-\frac{u^2(u+1)}{u+1}=u(u+1)-u^2=u[/tex]
In the second bracket let [tex]w=\sqrt[4]{a}[/tex] ([tex]\sqrt{a}=w^2[/tex]):
[tex]\frac{w+w^3}{w^2+1}+\frac{1-w^2}{w}=w+\frac{1-w^2}{w}=\frac{w^2+1-w^2}{w}=\frac{1}{w}[/tex]
[tex]u:\frac{1}{w}=uw=\sqrt[3]{a}\times \sqrt[4]{a}=a^{\frac{1}{3}+\frac{1}{4}}=\sqrt[12]{a^7}[/tex]