Elements of Algebra

382 ELEMENTS PART II. terms, being divided by $x^2$, give $3p^2+p^3x=1$, or $x=\frac{1-3p^2}{p^3}$. Now, $p=\frac{1}{3}$, so that $x=\frac{\frac{2}{3}}{\frac{1}{27}}=18$, and our formula becomes $1+18+324=343$, and the cube root $1+px=7$. If now we proceed, by making $x=18+y$, our formula will assume the form $343+37y+y^2$, and by the first rule we must suppose its cube root to be $7+py$; comparing it then with the cube, $343+147py+21p^2y^2+p^3y^3$, it is evident we must make $147p=37$, or $p=\frac{37}{147}$; the other terms give the equation $21p^2+py=1$, whence we obtain the value of $y=\frac{1-21p^2}{p^3}=-\frac{147\times(147^2-21\times37^2)}{37^3}=-\frac{1049580}{50653}$, which may lead, in the same manner, to new values. 154. Let it now be required to make the formula $2+x^2$ equal to a cube. Here, as we easily get the case $x=5$, we shall immediately make $x=5+y$, and shall have $27+10y+y^2=2+x^2$; supposing now its cube root to be $3+py$, so that the formula itself may be $27+27py+9p^2y^2+p^3y^3$, we shall have to make $27p=10$, or $p=\frac{10}{27}$; therefore $1=9p^2+p^3y$, and $y=\frac{1-9p^2}{p^3}=-\frac{27\times(27^2-9\times10^2)}{1000}=-\frac{4617}{1000}$, and $x=\frac{383}{1000}$; therefore our formula becomes $2+x^2=\frac{2146689}{1000000}$, the cube root of which must be $3+py=\frac{129}{100}$. 155. Let us also see whether the formula, $1+x^3$, can become a cube in any other cases beside the evident ones of $x=0$, and $x=-1$. We may here remark first, that though this formula belongs to the third class, yet the root $1+x$ is of no use to us, because its cube, $1+3x+3x^2+x^3$, being equal to the formula, gives $3x+3x^2=0$, or $3x(1+x)=0$, that is, again, $x=0$, or $x=-1$. If we made $x=-1+y$, we should have to transform into a cube the formula $3y-3y^2+y^3$, which belongs to the second case; so that, supposing its cube root to be $p+y$, or the formula itself equal to the cube, $p^3+3p^2y+3py^2+y^3$, we should have $3p=-3$, or $p=-1$, and thence the equation $3y=p^3+3p^2y=-1+3y$, which gives $y=\frac{1}{0}$, or infinity; so that we obtain nothing more from this second supposition. In fact, it is in vain to seek for other values of $x$; for it may be demonstrated, that the sum of two cubes, as $t^3+x^3$, can never become