[tex]\sqrt[3]{64}=4[/tex] so it's obvious, I guess :)
Now, let's prove [tex]\sqrt[3]{12}[/tex] is not rational number.
Proof by contradiction.
Let assume [tex]\sqrt[3]{12}[/tex] is a rational number. Therefore it can be expressed as a fraction [tex]\dfrac{a}{b}[/tex] where [tex]a,b\in\mathbb{Z}[/tex] and [tex]\text{gcd}(a,b)=1[/tex].
[tex]\sqrt[3]{12}=\dfrac{a}{b}\\\\12=\dfrac{a^3}{b^3}\\\\a^3=12b^3[/tex]
[tex]12b^3[/tex] is an even number, so [tex]a^3[/tex] must also be an even number, and therefore also [tex]a[/tex] must be an even number. So, we can say that [tex]a=2k[/tex], where [tex]k\in\mathbb{Z}[/tex].
[tex](2k)^3=12b^3\\\\8k^3=12b^3\\\\2k^3=3b^3[/tex]
Since [tex]2k^3[/tex] is an even number, then also [tex]3b^3[/tex] must be an even number. 3 is odd, so for [tex]3b^3[/tex] to be an even number, [tex]b^3[/tex] must be an even number, and therefore [tex]b[/tex] is an even number.
But if both a and b are even numbers, then it contradicts our earlier assumption that [tex]\text{gcd}(a,b)=1[/tex]. Therefore [tex]\sqrt[3]{12}[/tex] is not a rational number.
