The box has minimum surface area Show that a rectangular prism (box) of given volume has minimum surface area if the box is a cube. 
Could you give me some hints what we are supposed to do?? 
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EDIT: 
Having found that for $z=\frac{V}{xy}$ the function $A_{\star}(x, y)=A(x, y, \frac{V}{xy})$ has its minimum at $(\sqrt[3]{V}, \sqrt[3]{V})$, how do we conclude that the box is a cube?? 
We have that $x=y$. Shouldn't we have $x=y=z$ to have a cube?? 

Volume $V=xyz$ given. Area $A(x,y,z)=2(xy+yz+zx)$, to minimise, when $x,y,z>0$ and $xyz=V$.
Fact. If $a,b,c>0$, then $a+b+c\ge 3\sqrt[3]{abc}$, and equality holds if and only if $a=b=c$. 
Proof. We set $X=\sqrt[3]{a}$, $Y=\sqrt[3]{b}$ and $Z=\sqrt[3]{c}$. Then the identity
$$
X^3+Y^3+Z^3-3XYZ=\frac{1}{2}(X+Y+Z)\big((X-Y)^2+(Y-Z)^2+(Z-X)^2\big),\tag{1}
$$
holds. This means that 
$$
a+b+c-3\sqrt[3]{abc}=X^3+Y^3+Z^3-3XYZ\ge 0,
$$
as the right-hand-side of $(1)$ is non-negative, and the equality only if $X-Y=Y-Z=Z-X=0$ or $X=Y=Z=0$ or $a=b=c=0$. $\quad\Box$
Hence,
$$
\frac{A}{2}=xy+yz+zx\ge 3\sqrt[3]{xy\cdot yz\cdot zx}=3\sqrt[3]{x^2y^2z^2}=3V^{2/3}.
$$
and equality holds iff $xy=zx=yz$ or equivalently iff $x=y=z$.
Indeed, $A$ is minimised when $x=y=z$, and $A_{\mathrm{min}}=6V^{2/3}$.
