I can give you the general writing on what's written on the page for better understanding.
$$ PQ^2 = 2 BP \cdot QC $$
$$ (a + b - \sqrt{a^2 + b^2})^2 = 2(\sqrt{a^2 + b^2} - a)(\sqrt{a^2 + b^2} - b) $$
$$ \{ \sqrt[3]{(a+b)^2} - \sqrt[3]{a^2 - ab + b^2} \}^3 = 3(\sqrt[3]{a^3 + b^3} - a)(\sqrt[3]{a^3 + b^3} - b) $$
$$ \sqrt{A + B \sqrt[3]{p}} = \sqrt{\frac{B}{p + k^3}} \left( \frac{k^2}{2} + k \sqrt[3]{p} - \sqrt[3]{p^2} \right) $$
where $Bk^4 - 4Ak^3 - 8Bkp - 4Ap = 0$.
$F \frac{1 - \sqrt{1 - t^{24}}}{2} = e^{-\pi \sqrt{29}}$. then
$$ t^{24} + 9t^{20} + 5t^{16} - 2t^{12} - 5t^8 + 9t^4 - 1 = 0 $$
$$ \frac{t^6 + t^2}{1 - t^4} = \sqrt{\frac{\sqrt{29} - 5}{2}} $$
$$ \frac{t^3 + t \sqrt{\sqrt{29} - 2}}{1 + t^2 \sqrt{\sqrt{29} + 2}} = \sqrt[4]{\frac{\sqrt{29} - 5}{2}} $$
if $\sqrt[4]{1 - t^8} = t(1 + u^2)$, then $u^3 + u = \sqrt{2}$.
$F \cdot \frac{1 - \sqrt{1 - \frac{1}{64} t^{24}}}{2} = e^{-\pi \sqrt{79}}$. then
$$ t^5 - t^4 + t^3 - 2t^2 + 3t - 1 = 0 $$
$F \cdot \frac{1 - \sqrt{1 - \frac{1}{64} t^{24}}}{2} = e^{-\pi \sqrt{47}}$, then
$$ t^5 + 2t^4 + 2t^3 + t^2 - 1 = 0 $$
Edit
The substitution is a reverse-engineered parametrization of the relation
$$PQ^2 = 2 BP \cdot QC.$$
Let
$$s = \sqrt{a^2 + b^2}.$$
Then the proposed choice is
$$BP = s - a, \quad QC = s - b, \quad PQ = a + b - s.$$
Now check what happens:
$$PQ^2 = (a + b - s)^2$$
and
$$2 BP \cdot QC = 2(s - a)(s - b).$$
Expanding both sides gives
$$(a + b - s)^2 = a^2 + b^2 + 2ab + s^2 - 2s(a + b),$$
$$2(s - a)(s - b) = 2(s^2 - s(a + b) + ab).$$
These are equal exactly because $s^2 = a^2 + b^2$. So the identity is built in.
The idea is to write the three lengths from the same quantity $s$:
$$s - a, \quad s - b, \quad a + b - s.$$
That is a symmetric way to make a quadratic relation like
$$x^2 = 2yz$$
