Elements of Algebra

384 ELEMENTS PART II. and thence $15p^2 + p^3y = 1$, or $p^3y = 1 - 15p^2 = -\frac{109}{375}$; consequently, $y = -\frac{122625}{10648}$, and $x = -\frac{5497}{10648}$. Now, since $x$ may either be negative or positive, $x^3$ being found alone in the given formula, let us suppose $x = \frac{2+2y}{1-y}$, and our formula will become $\frac{8+8y^2}{(1-y)^2}$ which must be a cube; let us therefore multiply both terms by $1-y$, in order that the denominator may become a cube; and this will give $\frac{8-8y+8y^2-8y^3}{(1-y)^3}$: then we shall only have the numerator $8-8y+8y^2-8y^3$, or if we divide by $8$, only the formula $1-y+y^2-y^3$, to transform into a cube; which formula belongs to all the three cases. Let us, according to the first, take for the root $1-\frac{1}{3}y$; the cube of which is $1-y+\frac{1}{3}y^2-\frac{1}{27}y^3$; so that we have $1-y = \frac{1}{3}-\frac{1}{27}y$, or $27-27y=9-y$; therefore $y=\frac{9}{13}$; also, $1+y=\frac{22}{13}$, and $1-y=\frac{4}{13}$; whence $x=11$, as before. We should have exactly the same result, if we considered the formula as coming under the second case. Lastly, if we apply the third, and take $1-y$ for the root, the cube of which is $1-3y+3y^2-y^3$, we shall have $-1+y=-3+3y$, and $y=1$; so that $x=\frac{2}{0}$, or infinity; and, consequently, a result which is of no use. 159. But since we already know the two cases, $x=2$, and $x=11$, we may also make $x = \frac{2+11y}{1+y}$; for, by these means, if $y=0$, we have $x=2$; and if $y=\infty$, or infinity, we have $x=11$. Therefore, let $x = \frac{2+11y}{1+y}$, and our formula becomes $4 + \frac{4+44y+121y^2}{1+2y+y^2}$, or $\frac{8+52y+125y^2}{(1+y)^2}$. Multiply both terms by $1+y$, in order that the denominator may become a cube, and we shall only have the numerator, $8+60y+177y^2+125y^3$, to transform into a cube. And if, for this purpose, we suppose the root to be $2+5y$, we shall not only have the first terms disappear, but also the last. We may, therefore, refer our formula to the second case, taking $p+5y$ for the root, the cube of which is $p^3+15p^2y+75py^2+125y^3$; so that we must make $75p=177$, or $p=\frac{59}{25}$; and there will result $8+60y = p^3+15p^2y$, or $-\frac{2943}{125}y = \frac{80379}{15625}$, and $y=\frac{80379}{367875}$, whence we might obtain a value of $x$.