Calculating moment of inertia for a cuboid

Question

I am trying to calculate the moment of inertia tensor for a cuboid with half edges of width $a$, height $b$, depth, $c$ and mass, $m$, which according to wikipedia, is:

\begin{equation} I_c =\begin{bmatrix} \dfrac{1}{12}m(b^2 + c^2) & 0 & 0 \\ 0 & \dfrac{1}{12}m(a^2 + c^2) & 0 \\ 0 & 0 & \dfrac{1}{12}m(a^2 + b^2) \\ \end{bmatrix} \end{equation}

We are summing the square distances of the infinite point masses to each rotation axis, so we use Pythagoras' Theorem to find a parametric equation of the square distance of each point from each axis:

for x-axis: \begin{equation} d^2 = y^2 + z^2 \\ \end{equation} since, \begin{equation} r^2 = x^2 + y^2 + z^2 \\ \end{equation} \begin{equation} r^2 - x^2 = y^2 + z^2 \\ \end{equation} which also intuitively makes sense as we are taking the distance from the x-axis, and therefore doesn't have any x component. \begin{equation} \int_0^c \int_0^b \int_0^a (y^2 + z^2) \hspace{0.3em} dx \hspace{0.3em} dy \hspace{0.3em} dz \end{equation} \begin{equation} \int_0^c \int_0^b (y^2 + z^2)a \hspace{0.3em} dy \hspace{0.3em} dz \end{equation} \begin{equation} \int_0^c a\dfrac{1}{3}b^3 + abz^2 \hspace{0.3em} dz \end{equation} \begin{equation} ab\int_0^c \dfrac{1}{3}b^2 + z^2 \hspace{0.3em} dz \end{equation} \begin{equation} ab(\int_0^c \dfrac{1}{3}b^2 \hspace{0.3em} dz + \int_0^c z^2 \hspace{0.3em} dz) \end{equation} \begin{equation} \dfrac{1}{3}ab^3c + \dfrac{1}{3}abc^3 \end{equation}

\begin{equation} \dfrac{1}{3}abc(b^2 + c^2) \end{equation}

similary for y-axis: \begin{equation} \dfrac{1}{3}abc(a^2 + c^2) \end{equation}

and z-axis: \begin{equation} \dfrac{1}{3}abc(a^2 + b^2) \end{equation}

we multiply each by $m$ by the definition of the inertia, $I = mr^2$, and align along the diagonal of the matrix.

\begin{equation} I_c =\begin{bmatrix} \dfrac{1}{3}m(b^2 + c^2) & 0 & 0 \\ 0 & \dfrac{1}{3}m(a^2 + c^2) & 0 \\ 0 & 0 & \dfrac{1}{3}m(a^2 + b^2) \\ \end{bmatrix} \end{equation}

Can anyone see where I went wrong?

My solution seems to be off by a scalar of $\dfrac{1}{4}$ for each axis.

TIA