Transform one-pole cascaded system into a biquadratic section I have two cascaded one-pole sections with the same coefficient b for both sections.
$$
y_A[n]=x_A[n]+b\cdot(y_A[n−1]−x_A[n])\\
y_B[n]=y_A[n]+b\cdot(y_B[n−1]−y_A[n])
$$

The final output of the system is $y_B[n]$.

I would like to re-implement the system as one biquad section of the form:

$$
y[n]=a_0x[n]+a_1x[n-1]+a_2x[n-2]-b_1y[n-1]-b_2y[n-2]
$$

The final output of the system is $y[n]$.

I would like to know what calculations I have to perform in order to make the output of the biquad exactly match the output of the one-pole cascaded system.
$$
y_B[n] = y[n]
$$
Obviously the 5 biquad coefficients $a_0\; a_1\; a_2\; b_1\; b_2$ will have to be determined as a function of the $b$ coefficient as found in the cascaded system above.

Can someone lead me through the math?

Thank you.
A single one-pole filter satisfies the following difference equation:

$$y[n]=x[n]+b\left(y[n-1]-x[n]\right)=(1-b)x[n]+by[n-1]\tag{1}$$

which corresponds to this transfer function:

$$H(z)=\frac{1-b}{1-bz^{-1}}\tag{2}$$

If you cascade two of these systems, the transfer functions are multiplied:

$$H^2(z)=\frac{(1-b)^2}{(1-bz^{-1})^2}=\frac{(1-b)^2}{1-2bz^{-1}+b^2z^{-2}}\tag{3}$$

The transfer function in $(3)$ corresponds to the following difference equation for the cascaded system:

$$y[n]=(1-b)^2x[n]+2by[n-1]-b^2y[n-2]\tag{4}$$

Consequently, implementing $(4)$ is equivalent to implementing $(1)$ twice.