How to solve that? I have no idea about how to solve the following:
$$\sqrt[4]{13x+1} + \sqrt[4]{4x-1} = 3\sqrt[4]{x}$$
Could somebody help me, please?

Let $\sqrt[4]{13x+1}=a$ and $ \sqrt[4]{4x-1}=b$.
Thus, $a\geq0$ and $b\geq0$ and $$a+b=3\sqrt[4]{\frac{a^4+b^4}{17}},$$ which is homogeneous and
$$\frac{a^4-1}{13}=\frac{b^4+1}{4}.$$
The first gives after squaring of the both sides:
$$(a-2b)(2a-b)(16a^2+23ab+16b^2)=0.$$
Since $16a^2+23ab+16b^2=0$ gives $a=b=0,$ which is impossible, 
we get $(a-2b)(2a-b)=0$ only. I think now you can end it.
