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

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Make $$ax^2+bx^2$$ the first Term of $$2p$$, then will $$\frac{1}{2}ax^2+\frac{1}{2}bx^2$$ be the firt Term of $$p$$. Therefore $$-abx^4-b^2x^4$$ will be the firt Term of $$2bx^{2}p$$, and $$\frac{1}{4}abx^4+\frac{1}{2}abx^4+\frac{1}{4}bx^4$$ will be the firt Term of $$p^2$$. Thee being collected, and their Signs changed, mut be made the econd Term of $$2p$$, which will give $$\frac{1}{4}abx^4+\frac{3}{8}b^2x^4 - \frac{1}{8}a^2x^4$$ for the econd Term of $$p$$. Then the econd 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 econd Term of $$p^2$$ (by quaring) 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 firt 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 o on as far as you pleae.

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 Subitution $$a^{3} + 3a^{2}p + 3ap^{2} + p^{3} = a^{3} + x^{3}, or 3a^{2}p+3ap^{2}+p^{3}=x^{3}$$, which upplemental Equation may be thus reolved.

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