<s>[INST] How to solve this equation: $\frac{\sqrt[5]{x^3 \sqrt{x\sqrt[3]{x^{-2}}}}}{\sqrt[4]{x\sqrt[3]{x}}}=3$ Please, help me to solve this equation:
$$\frac{\sqrt[5]{x^3\sqrt{x\sqrt[3]{x^{-2}}}}}{\sqrt[4]{x\sqrt[3]{x}}}=3$$
I tried to shorten fraction, but I get very weird numbers like 
$$\frac{\sqrt[30]{x^{19}}}{\sqrt[3]{x}}=3,$$ and I'm stuck there :(
 [/INST] If $x\gt 0$, then:


*

*$\sqrt[3]{x^{-2}} = x^{-2/3}$.

*$x\sqrt[3]{x^{-2}} = xx^{-2/3} = x^{1/3}$.

*$\sqrt{x\sqrt[3]{x^{-2}}} = x^{1/6}$.

*$x^3\sqrt{x\sqrt[3]{x^{-2}}} = x^{19/6}$.

*$\sqrt[5]{x^3\sqrt{x\sqrt[3]{x^{-2}}}} = x^{19/30}$.

*$x\sqrt[3]{x} = xx^{1/3} = x^{4/3}$.

*$\sqrt[4]{x\sqrt[3]{x}} = x^{1/3}$.


So the quotient is equal to $x^{19/30}/x^{1/3} = x^{(19/30)-(1/3)} = x^{9/30}$. Your equation is then equivalent to
$$x^{9/30} = 3$$
which can be solved by raising both sides to the $30/9$.
If $x\lt 0$, then you need to throw in a few absolute values, since for example, $\sqrt{x\sqrt[3]{x^{-2}}} = |x|^{1/6}$, instead of $x^{1/6}$. 
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