+118654 (-81)^\frac{1}{4}
Firstly there is no real solution for this because no real number to the power of 4 is going to be negative.
But I will give you the complex solution :)
\((-81)^\frac{1}{4}\\ =(-1)^\frac{1}{4}*(81)^\frac{1}{4}\\ =(-1)^\frac{1}{4}*3\\ =((-1)^\frac{1}{2})^\frac{1}{2}*3\\ =i^\frac{1}{2}*3\)
Now
\(\sqrt{i}=\sqrt{\frac{1+2i-1}{2}}=\sqrt{\frac{1+2i+i^2}{2}}=\sqrt{\frac{(1+i)^2}{2}}=\frac{1+i}{\sqrt2}\\ \)
\(=3*\sqrt{i}\\ =\frac{3(1+i)}{\sqrt2}\\ =\frac{3\sqrt2 (1+i)}{2}\\\)
+118654 I forgot to finish it............this is one answer but there are 3 more...
I see Heureka is answering so I expect he will give the other ones. :)
Thanks in advance Heureka 
+26388 Hello Melody,
you set -1 to \(i^2\) but i think this is not allowed, if (-1) is a complex number.
only (-1) real is i^2 and (-1) complex is \(e^{i\pi}\).
This way is okay \(i^2 \rightarrow -1\)
This way is not allowed \(-1 \rightarrow i^2\)
+118654 Hi Heureka,
I certainly accept that your knowledge of mathematics is far superior to mine but I do not get why I cannot do this.
Our answers are the same.
I suppose it is some mathematics nuance that has gone over my head :(
But thank you 
+26388 (-81)^\frac{1}{4}
\(\begin{array}{|rcll|} \hline x&=& \sqrt[4]{-81} \\ &=& (-81)^\frac{1}{4} \\ &=& (81)^\frac{1}{4} \cdot (-1)^\frac{1}{4} \\ &=& 3 \cdot (-1)^\frac{1}{4} \quad & | \quad e^{i\pi} = -1\ ! \quad (\text{Euler's Formula for Complex Numbers}) \\ x_1 &=& 3\cdot (e^{i\pi})^\frac{1}{4} \\ &=& 3\cdot e^{i\frac{\pi}{4}} \\ &=& 3\cdot [\cos(\frac{\pi}{4}) + i\cdot \sin(\frac{\pi}{4})] \quad & | \quad \cos(\frac{\pi}{4})=\sin(\frac{\pi}{4})= \frac{\sqrt{2}}{2} \\ &=& 3\cdot (\frac{\sqrt{2}}{2}+ i\cdot \frac{\sqrt{2}}{2} ) \\ &=& \frac{3 \sqrt{2}}{2}(1+i) \\\\ x_2 &=& 3\cdot e^{i(\frac{\pi}{4}+\frac{\pi}{2} )} \\ x_2 &=& 3\cdot e^{\frac{3}{4} i \pi } \\\\ x_3 &=& 3\cdot e^{i(\frac{\pi}{4}+2\cdot\frac{\pi}{2} )} \\ x_3 &=& 3\cdot e^{\frac{5}{4} i \pi } \\\\ x_4 &=& 3\cdot e^{i(\frac{\pi}{4}+3\cdot\frac{\pi}{2} )} \\ x_4 &=& 3\cdot e^{\frac{7}{4} i \pi } \\\\ \hline \end{array} \)
