<s>[INST] How to find the value of $x$ in $x^5=32$ I understand that $2^5=32$ but how would one go about finding it without doing any guessing (what if the numbers were much greater)?
 [/INST] USING COMPLEX ANALYSIS
We have 
$$z^5=2^5\implies z=2e^{i2n\pi/5}$$
for $n=0, \pm 1, \pm 2$.  
$$\bbox[5px,border:2px solid #C0A000]{\text{Thus the five roots are} \,\, 2,\, 2e^{\pm i2\pi/5},\,2e^{\pm i4\pi/5}}\tag 1$$

FACTORORING A POLYNOMIAL
We have $x^5=32\implies x^5-2^5=0$.  
We can then factor $x-2$ from the left-hand side and write 
$$x^5-2^5=(x-2)(x^4+2x^3+4x^2+8x+16)=0 \tag 1$$
The quartic expression in $(1)$ can be factored into the product of two quadratics as 
$$x^4+2x^3+4x^2+8x+16=(x^2+ax+4)(x^2+bx+4) \tag 2$$
where by matching coefficients in $(2)$, we find $a+b=2$ and $ab=-4$.  
Solving for $a$ and $b$ reveals that $a=1+\sqrt{5}$ and $b=1-\sqrt{5}$ whence the original polynomial can be written as
$$x^5-2^5=(x-2)(x^2+(1+\sqrt{5})x+4)(x^2+(1-\sqrt{5})x+4) \tag 3$$
Finally, the quadratic terms in $(3)$ can easily be factored as 
$$x^2+(1+ \sqrt{5})x+4=\left(x-\frac{(1+\sqrt{5})+i\sqrt{10-2\sqrt{5}}}{2}\right)\left(x-\frac{(1+\sqrt{5})-i\sqrt{10-2\sqrt{5}}}{2}\right)$$
and
$$x^2+(1- \sqrt{5})x+4=\left(x-\frac{(1-\sqrt{5})+i\sqrt{10+2\sqrt{5}}}{2}\right)\left(x-\frac{(1-\sqrt{5})-\sqrt{10+2\sqrt{5}}}{2}\right)$$
Thus the five roots of $x^5-2^5=0$ are
$$\bbox[5px,border:2px solid #C0A000]{2,\, \frac{(1+  \sqrt{5})\pm i\sqrt{10- 2\sqrt{5}}}{2},\,\frac{(1-  \sqrt{5})\pm i\sqrt{10+ 2\sqrt{5}}}{2}} \tag 4$$
where the roots in $(4)$ are the rectangular coordinate form of the polar roots in $(1)$.
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