For your apparent purpose, in dimension $n$ it is convenient to begin with a homogeneous polynomial of total degree $2n$ with all individual exponents even. Then, for translates and rotations, all sorts of lower degree and odd exponent terms may show up.
In $\mathbb R^2,$ a rounded version of an ordinary square is $$A(x^4 + y^4) + B x^2 y^2 = 1.$$ The ordinary unit circle is $A=1, B = 2.$ Disjoint hyperbolas are $A=1, B=-2.$ A somewhat squared shape, indeed the $L^4$ "unit circle," is $A=1, B=0.$ An alternative to Piet Hein's "superellipse" is $A=1, B=1.$ Finally, a real analytic curve that passes through all 8 lattice points with $|x| \leq 1,\; |y| \leq 1$ other than the origin itself is $A=1, B = -1.$1,$or $$x^4 - x^2 y^2 + y^4 = 1.$$ At some point I wanted a smooth version of an ordinary cube in$\mathbb R^3,$meaning that it passed through all 26 integer lattice points with$ |x| \leq 1,\; |y| \leq 1,\; |z| \leq 1$other than the origin itself. I wrote $$A( x^6 + y^6 + z^6) + B (y^4 z^2 + z^4 x^2 + x^4 y^2 + y^2 z^4 + z^2 x^4 + x^2 y^4) + C x^2 y^2 z^2 = 1.$$ To find$A,B,C$it is only necessary to check the$(x,y,z)$triples$(0,0,1),(0,1,1),(1,1,1),$and evidently$A=1, B=-1B=-\frac{1}{2}, C=1$works. so $$( x^6 + y^6 + z^6) - (y^4 \frac{1}{2}(y^4 z^2 + z^4 x^2 + x^4 y^2 + y^2 z^4 + z^2 x^4 + x^2 y^4) + x^2 y^2 z^2 = 1$$ is a rounded cube. I recall graphing this in spherical coordinates with$\rho$a function of$\theta, \phi.$The trouble was that it is very flat near the axes, so without spherical coordinates many different patches were necessary. It is obvious that this is a star-shaped body around the origin, it needs just a little more work to confirm that it is compact. 1 For your apparent purpose, in dimension$n$it is convenient to begin with a homogeneous polynomial of total degree$2n$with all individual exponents even. Then, for translates and rotations, all sorts of lower degree and odd exponent terms may show up. In$\mathbb R^2,$a rounded version of an ordinary square is $$A(x^4 + y^4) + B x^2 y^2 = 1.$$ The ordinary unit circle is$A=1, B = 2.$Disjoint hyperbolas are$A=1, B=-2.$A somewhat squared shape, indeed the$L^4$"unit circle," is$A=1, B=0.$An alternative to Piet Hein's "superellipse" is$A=1, B=1.$Finally, a real analytic curve that passes through all 8 lattice points with$ |x| \leq 1,\; |y| \leq 1$other than the origin itself is$A=1, B = -1.$At some point I wanted a smooth version of an ordinary cube in$\mathbb R^3,$meaning that it passed through all 26 integer lattice points with$ |x| \leq 1,\; |y| \leq 1,\; |z| \leq 1$other than the origin itself. I wrote $$A( x^6 + y^6 + z^6) + B (y^4 z^2 + z^4 x^2 + x^4 y^2) + C x^2 y^2 z^2 = 1.$$ To find$A,B,C$it is only necessary to check the$(x,y,z)$triples$(0,0,1),(0,1,1),(1,1,1),$and evidently$A=1, B=-1, C=1$works. so $$( x^6 + y^6 + z^6) - (y^4 z^2 + z^4 x^2 + x^4 y^2) + x^2 y^2 z^2 = 1$$ is a rounded cube. I recall graphing this in spherical coordinates with$\rho$a function of$\theta, \phi.\$ The trouble was that it is very flat near the axes, so without spherical coordinates many different patches were necessary. It is obvious that this is a star-shaped body around the origin, it needs just a little more work to confirm that it is compact.