How Einstein missed his opportunity to derive Lorentz in Point §3.

I was reading again the English version (1923) of his 1905 paper, and
got
interested in a footnote from the editor:
QUOTE:
5The equations of the Lorentz transformation may be more simply deduced
directly from the condition that in virtue of those equations the
relation
x² + y² + z²  = c² t² shall have as its consequence the second relation
ξ² + η² + ζ²  = c²τ².
END QUOTE
As the above equations are EXACTLY what Poincaré postulated about space
and time,
and because Poincaré and Minkowski knew each other, so the latter used
Poincaré to
derive the spacetime shit, I searched about this kind of derivation
without Taylor, mirrors and CHEATS/HACKS.
This is based on the footnote and what Einstein wrote at the end of
Point §3 (PLAGIARIZED EXACTLY FROM POINCARÉ, SAME YEAR).
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QUOTING EINSTEIN IN Point §3
--------------------------------------------------------------------------
We now have to prove that any ray of light, measured in the moving
system,
is propagated with the velocity c, if, as we have assumed, this is the
case
in the stationary system; for we have not as yet furnished the proof
that
the principle of the constancy of the velocity of light is compatible
with
the principle of relativity.
At the time t = τ = 0, when the origin of the co-ordinates is common to
the
two systems, let a spherical wave be emitted therefrom, and be
propagated
with the velocity c in system K. If (x, y, z) be a point just attained
by
this wave, then
x² + y² + z²  = c² t²
Transforming this equation with the aid of our equations of
transformation
we obtain after a simple calculation
ξ² + η² + ζ²  = c² τ²
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This is the derivation of Lorentz transforms using the above concepts:
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He could have derived Lorentz transforms by simply postulating that as
c² t² - x² - y² - z²  = 0
and
c² τ² - ξ² - η² - ζ²  = 0
then
c² τ² - ξ² - η² - ζ²  = c² t² - x² - y² - z²
As y = η  and  z = ζ, this equation reduces to c² τ² - ξ²  = c² t² - x²
Assuming that ξ and x differs in a factor β from Galilean transform,
where β = 1, the new transform can be written as
ξ  = β (x - vt)
and
x  = β (ξ + v τ)
then
x  = β² (x - vt) + β v τ
β v τ = x (1 - β²) + β² v t
τ = x (1 - β²)/βv + β  t
τ = β [(1/β² - 1) x/v +  t]
Replacing ξ and τ in ξ² = c² τ², it follows that
β² (x - vt)² = c² β² [x (1/β² - 1) x/v +  t]²
Factoring the above equation with a little help of algebra, it's
obtained
[β² - c²β²/v² (1/β² - 1)] x² - 2 β² [v + c²/v (1/β² - 1)] xt = β² (c² -
v²) t²
To verify
x² + y² + z²  = c² t²
It's required that, in the previous equation,
[β² - c²β²/v² (1/β² - 1)] = 1
[v + c²/v (1/β² - 1)] = 0
β² (c² - v²) = c²
From the last equation,
β²  = c²/(c² - v²)
or
β  = 1/√(1 - v²/c²) , which is the modern factor Gamma (γ).
With this result of β  (γ), the transforms are
ξ  = β (x - vt) = (x - vt)/√(1 - v²/c²)
τ = β [(1/β² - 1) x/v +  t] = β (t - vx/c²) = (t - vx/c²)/√(1 - v²/c²)
Without the complex and FRAUDULENT use of calculus and algebra in the
first part of Point §3.
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