Let's assume the cube has corners $(±1,±1,±1)$ and you have $P_1:x=y$ and $P_2:x=z$. So the common corners are $(1,1,1)$ and $(-1,-1,-1)$.
One of the pieces has $x<y\wedge x<z$. Its volume is therefore
$$\int_{-1}^1\mathrm dx\int_x^1\mathrm dy\int_x^1\mathrm dz=
\int_{-1}^1(1-x)^2\,\mathrm dx=\left[\frac13x^3-x^2+x\right]_{-1}^1=\frac83$$
Another piece has $y<x\wedge x<z$. Its volume is
$$\int_{-1}^1\mathrm dx\int_{-1}^x\mathrm dy\int_x^1\mathrm dz=
\int_{-1}^1(1+x)(1-x)\,\mathrm dx=\left[x-\frac13x^3\right]_{-1}^1=\frac43$$
The piece $x>y\wedge x>z$ is just like $x<y\wedge x<z$ so it has volume $\frac83$. The piece $z<x\wedge x<y$ is just like $y<x\wedge x<z$ with volume $\frac43$. Together these pieces have volume $2\left(\frac83+\frac43\right)=8=2^3$ which matches the volume of the whole cube. If your cube has a different edge length than $2$, scale these figures accordingly.