Prove that $\sqrt[3]{6}$ is an irrational number.

(N/A) Assume,for the sake of contradiction,that $\sqrt[3]{6}$ is a rational number.
Let $\sqrt[3]{6} = \frac{a}{b}$,where $a, b \in N$ and $g.c.d.(a, b) = 1$.
Since $1 < 6 < 8$,we have $\sqrt[3]{1} < \sqrt[3]{6} < \sqrt[3]{8}$,which implies $1 < \frac{a}{b} < 2$.
This means $b > 1$,because if $b = 1$,then $\frac{a}{b}$ would be an integer,but there is no integer between $1$ and $2$.
Cubing both sides,we get $6 = \frac{a^3}{b^3}$,which implies $6b^3 = a^3$.
Since $g.c.d.(a, b) = 1$,it follows that $g.c.d.(a^3, b^3) = 1$.
From $6b^3 = a^3$,we see that $b^3$ must divide $a^3$. Since $g.c.d.(a^3, b^3) = 1$,this is only possible if $b^3 = 1$,which means $b = 1$.
However,we already established that $b > 1$. This is a contradiction.
Therefore,our assumption that $\sqrt[3]{6}$ is rational is false.
Hence,$\sqrt[3]{6}$ is an irrational number.