# Page:The method of fluxions and infinite series.djvu/209

Make ${\displaystyle ax^{2}+bx^{2}}$ the first Term of ${\displaystyle 2p}$, then will ${\displaystyle {\frac {1}{2}}ax^{2}+{\frac {1}{2}}bx^{2}}$ be the firſt Term of ${\displaystyle p}$. Therefore ${\displaystyle -abx^{4}-b^{2}x^{4}}$ will be the firſt Term of ${\displaystyle 2bx^{2}p}$, and ${\displaystyle {\frac {1}{4}}abx^{4}+{\frac {1}{2}}abx^{4}+{\frac {1}{4}}bx^{4}}$ will be the firſt Term of ${\displaystyle p^{2}}$. Theſe being collected, and their Signs changed, muſt be made the ſecond Term of ${\displaystyle 2p}$, which will give ${\displaystyle {\frac {1}{4}}abx^{4}+{\frac {3}{8}}b^{2}x^{4}-{\frac {1}{8}}a^{2}x^{4}}$ for the ſecond Term of ${\displaystyle p}$. Then the ſecond Term of ${\displaystyle -2bx^{2}p}$ will be ${\displaystyle -{\frac {1}{2}}ab^{2}x^{6}-{\frac {3}{4}}x^{6}+{\frac {1}{4}}a^{2}bx^{6}}$ and the ſecond Term of ${\displaystyle p^{2}}$ (by ſquaring) will be found ${\displaystyle {\frac {1}{8}}a^{2}bx^{6}+{\frac {5}{8}}ab^{2}x^{6}-{\frac {1}{8}}a^{2}x^{6}+{\frac {3}{8}}b^{3}x^{6}}$ and the firſt Term of ${\displaystyle -bx^{2}p^{2}}$ will be ${\displaystyle -{\frac {1}{4}}a^{2}bx^{6}-{\frac {1}{2}}ab^{2}x^{6}-{\frac {1}{4}}b^{3}x^{6};}$ which being collected and the Signs changed, will make the third Term of ${\displaystyle 2p}$, half which will be the third Term of ${\displaystyle p}$; and ſo on as far as you pleaſe.
And thus if we were to extract the Cube-root of ${\displaystyle a^{3}+x^{3}}$, or the Root ${\displaystyle y}$ of this Equation ${\displaystyle y^{3}=a^{3}+x^{3}}$; make ${\displaystyle y=a+p}$, then by Subſitution ${\displaystyle a^{3}+3a^{2}p+3ap^{2}+p^{3}=a^{3}+x^{3},or3a^{2}p+3ap^{2}+p^{3}=x^{3}}$ , which ſupplemental Equation may be thus reſolved.