Let $\vartheta=\sqrt[3]{175}$, and let $K=\mathbb{Q}(\vartheta)$. Then it can be shown that $1, \vartheta, \frac{\vartheta^2}{5}$ is an integral basis for $K$. This should prompt you towards the discovery that $\sqrt[3]{245}=\frac{\vartheta^2}{5}$, and hence $\alpha=\frac{1}{3}(x+y\vartheta+\frac{z}{5}\vartheta^2)$.
So you would require $3\mid x,y,z$ if $\alpha$ was to be an algebraic integer. Given that $0\le x,y,z\le 2$, this can only happen if $x=y=z=0$.
Edit. Again let $K=\mathbb{Q}(\vartheta)\cong\mathbb{Q}[x]/(x^3-175)=\mathbb{Q}[x]/(f)$. Recall that $\Delta(\vartheta)=[\mathcal{O}_K:\mathbb{Z}[\vartheta]]^2\Delta_K$, and $\Delta(\vartheta)=-27(-175)^2=-826875=-3^3\cdot 5^4\cdot 7^2$. So the only rational primes that could divide the index are $3$, $5$, and $7$, and we can factor $(3)$, $(5)$, and $(7)$ to see whether or not they do..
($p=3$) Then $f(x)=(x-2)^3$ modulo $3$. Since $3^2\nmid f(2)$, $(3)=\mathfrak{p}_3^3$, where $\mathfrak{p}_3=(3,\vartheta-2)$ is invertible: so $3$ does not divide the index.
($p=7$) Then $f(x)=x^3$ modulo $7$. Since $7^2\nmid f(0)$, $(7)=\mathfrak{p}_7^3$, where $\mathfrak{p}_7=(7,\vartheta)$ is invertible: so $7$ does not divide the index.
($p=5$) Then $f(x)=x^3$ again modulo $5$. But $5^2\mid f(0)$, so we know that we need to enlarge our ring, and we can take an element in $\mathcal{O}_K\setminus\mathbb{Z}[\vartheta]$. When dividing $f(x)$ by $x$ (it's irreducible factor modulo $5$), we get $f(x)=\color{red}{x^2}\cdot x-175$, hence $\frac{\vartheta^2}{5}\in\mathcal{O}_K\setminus\mathbb{Z}[\alpha]$.
Now the minimal polynomial of $\frac{\vartheta^2}{5}$ is $g(x)=x^3-245$, and $g(x)=x^3$ modulo $5$. Since $5^2\nmid g(0)$, we know that this is the only element we need to add to $\mathbb{Z}[\vartheta]$, and $\mathcal{O}_K=\mathbb{Z}[\vartheta, \frac{\vartheta^2}{5}]$.
