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

Make $ax^{2}+bx^{2}$ the first Term of $2p$, then will ${\frac {1}{2}}ax^{2}+{\frac {1}{2}}bx^{2}$ be the firſt Term of $p$. Therefore $-abx^{4}-b^{2}x^{4}$ will be the firſt Term of $2bx^{2}p$, and ${\frac {1}{4}}abx^{4}+{\frac {1}{2}}abx^{4}+{\frac {1}{4}}bx^{4}$ will be the firſt Term of $p^{2}$. Theſe being collected, and their Signs changed, muſt be made the ſecond Term of $2p$, which will give ${\frac {1}{4}}abx^{4}+{\frac {3}{8}}b^{2}x^{4}-{\frac {1}{8}}a^{2}x^{4}$ for the ſecond Term of $p$. Then the ſecond Term of $-2bx^{2}p$ will be $-{\frac {1}{2}}ab^{2}x^{6}-{\frac {3}{4}}x^{6}+{\frac {1}{4}}a^{2}bx^{6}$ and the ſecond Term of $p^{2}$ (by ſquaring) will be found ${\frac {1}{8}}a^{2}bx^{6}+{\frac {5}{8}}ab^{2}x^{6}-{\frac {1}{8}}a^{2}x^{6}+{\frac {3}{8}}b^{3}x^{6}$ and the firſt Term of $-bx^{2}p^{2}$ will be $-{\frac {1}{4}}a^{2}bx^{6}-{\frac {1}{2}}ab^{2}x^{6}-{\frac {1}{4}}b^{3}x^{6};$ which being collected and the Signs changed, will make the third Term of $2p$, half which will be the third Term of $p$; and ſo on as far as you pleaſe.

And thus if we were to extract the Cube-root of $a^{3}+x^{3}$, or the Root $y$ of this Equation $y^{3}=a^{3}+x^{3}$; make $y=a+p$, then by Subſitution $a^{3}+3a^{2}p+3ap^{2}+p^{3}=a^{3}+x^{3},or3a^{2}p+3ap^{2}+p^{3}=x^{3}$ , which ſupplemental Equation may be thus reſolved.

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