Step 1: Write the equations clearly.
The three equations are $3x+y+2z=-2$, $2x-3y+z=7$, and $x-4y+3z=1$. We will solve step by step using elimination.
Step 2: Remove $x$ from two equations.
Double the third equation: $2x-8y+6z=2$. Subtract the second from it: $(2x-8y+6z)-(2x-3y+z)=2-7$, giving $-5y+5z=-5$, so $z-y=-1$ and $z=y-1$.
Step 3: Use this in the second equation.
Put $z=y-1$ into $2x-3y+z=7$: $2x-3y+y-1=7$, so $2x-2y=8$, which means $x-y=4$ and $x=y+4$.
Step 4: Use both in the first equation.
Put $x=y+4$ and $z=y-1$ into $3x+y+2z=-2$: $3(y+4)+y+2(y-1)=-2$. This becomes $6y+10=-2$.
Step 5: Find each value.
So $6y=-12$, giving $y=-2$. Then $x=2$ and $z=-3$. Thus $\alpha=2$ and $\beta=-2$.
Step 6: Compute the asked expression.
Finally $\alpha^3-\beta^3=2^3-(-2)^3=8+8=16$. \[ \boxed{16} \]