Determine a number is transcendental/algebraic 
Determine: $(0.064)^{\frac{1}{3}}$ is transcendental or algebraic 

To show a number is transcendental/algebraic do I need to show there is a monic polynomial with integer coefficients such that the number is/is not its root ?
in this case:
$x=(0.064)^{\frac{1}{3}}\iff x=\frac{2}{5}\iff x-\frac{2}{5}=0$
How can I continue?
 A: Either a monic polynomial with rational coefficients, or any polynomial with integer coefficients.
For example $5x^3 -29x^2+3x+11 = 0$ has integer coefficients, and is equivalent to $x^3 - \frac{29}5 x^2 + \frac 3 5 x + \frac{11}5 =0\vphantom{\dfrac11}$, which is monic and has rational coefficients.
For that matter, a non-monic polynomial with rational coefficients can serve.  For example if you have $\frac 7 6 x^3 + \frac 3 8 x^2 + 5x + \frac 1 2 = 0\vphantom{\dfrac11}$ you observe that the smallest common multiple of the denominators $6$, $8$, and $2$ is $24$, and multiplying both sides by that, you get $28x^3 + 9x^2 +120x + 12 =0\vphantom{\dfrac11}$, and then if you want it to be monic, just divide every term on both sides by $28$. 
A: https://en.wikipedia.org/wiki/Algebraic_number
A number is algebraic if it is a root of a non-zero polynomial with rational coefficients. There is no demand that the polynomial be monic.
A: A number $y$ is algebraic if it satisfies a polynomial $p(\alpha)$ having rational numbers as coefficients; that is, if $p(y) = 0$ where $p(\alpha) \in \Bbb Q[\alpha]$.  Note we may in fact assume $p(\alpha)$ is monic, since $p(y) = 0$ if and only if $\lambda^{-1} p(y) = 0$, where $\lambda \in \Bbb Q$ is the leading coefficient of $p(\alpha)$; clearly $rp(\alpha) \in \Bbb Q[\alpha]$ for any $r \in \Bbb Q$.
These things being the case, in the present instance we have
$x = (.064)^{1/3}, \tag{1}$
$x^3 =.064, \tag{2}$
whence
$x^3 - .064 = 0; \tag{3}$
note that
$.064 = \dfrac{64}{1000} = \dfrac{16}{250}$
$= \dfrac{8}{125} \in \Bbb Q; \tag{4}$
thus
$x^3 - \dfrac{8}{125} = 0; \tag{5}$
$x$ is therefore algebraic over $\Bbb Q$, satisfying as it does the cubic polynomial
$\alpha^3 - \dfrac{8}{125} \in \Bbb Q[\alpha]. \tag{6}$
