Newton's Method of Fluxions pg.8 paragraph $25$ question y'all.  Newton has this problem.  I'm using the free copy downloaded from the Library of Congress website, the original English.  In paragraph 25 line 7 he states $y \times b - x = v$ . . . by which I guess he means $y(b-x)$.  Then he says to write,$ v/(b-x) = y$.  Then in the next line he gets that long equation:
$$v^3 + b^2v^2 - b^3x^3 + 3b^2x^4 - 3bx^5 + x^6 = 0$$
How the heck did he get this?  Could it be a mistranslation from the Latin?  Thanks.
Quote:

Source:
The method of fluxions and infinite series; with its application to the geometry of curve-lines by Sir Isaac Newton, published in 1736, p.8

No, there is no mistake. He has the equation $y^3+\frac{b^2}{b-x}\cdot y^2-x^3=0$. At the next step we substitute $y$ by $\frac{v}{b-x}$
$$\left(\frac{v}{b-x}\right)^3+\frac{b^2}{b-x}\cdot \left(\frac{v}{b-x}\right)^2-x^3=0$$
$$\frac{v^3}{(b-x)^3}+\frac{b^2v^2}{(b-x)^3}-x^3=0$$
Expanding the last summand by $(b-x)^3$.
$$\frac{v^3}{(b-x)^3}+\frac{b^2v^2}{(b-x)^3}-\frac{x^3\cdot (b-x)^3}{(b-x)^3}=0$$
We can ommit the denominator, since the numerator has to be 0 only.
$$ v^3+b^2v^2-x^3\cdot (b-x)^3=0$$
To  expand $(b-x)^3$ you can use  the binomial theorem: $(b-x)^3=\sum\limits_{k=0}^3 \binom{3}{k}\cdot b^k\cdot (-x)^{3-k}$
$$v^3+b^2v^2-x^3b^3+3x^4b^2-3x^5b+x^6=0$$
