Elements of Algebra

CHAP. X. OF ALGEBRA. 383 a cube*; so that, by making $t=1$, it follows that the formula, $a^3+1$, can never become a cube, except in the cases already mentioned. 156. In the same manner, we shall find that the formula, $x^3+2$, can only become a cube in the case of $x=-1$. This formula belongs to the second case; but the rule there given cannot be applied to it, because the middle terms are wanting. It is by supposing $x=-1+y$, which gives $1+3y-3y^2+y^3$, that the formula may be managed according to all the three cases, and that the truth of what we have advanced may be demonstrated. If, in the first case, we make the root $=1+y$, whose cube is $1+3y+3y^2+y^3$, we have $-3y^2=3y^2$, which can only be true when $y=0$: and if, according to the second case, the root be $-1+y$, or the formula equal to $-1+3y-3y^2+y^3$, we have $1+3y=-1+3y$, and $y=\frac{2}{0}$, or an infinite value; lastly, the third case requires us to suppose the root to be $1+y$, which has already been done for the first case. 157. Let the formula $3x^3+3$ be also required to be transformed into a cube. This may be done, in the first place, if $x=-1$; but from that we can conclude nothing: then also, when $x=2$; and if, in this second case, we suppose $x=2+y$, we shall have the formula $27+36y+18y^2+3y^3$; and as this belongs to the first case, we shall represent its root by $3+py$, the cube of which is $27+27py+9p^2y^2+p^3y^3$; then, by comparison, we find $27p=36$, or $p=\frac{4}{3}$; and thence results the equation, $18+3y=9p^2+p^3y=16+\frac{64}{27}y$; which gives $y=\frac{-54}{17}$, and, consequently, $x=\frac{-20}{17}$: therefore our formula $3+3x^3 = -\frac{9261}{4913}$, and its cube root $3+py=\frac{21}{17}$; which solution would furnish new values, if we chose to proceed. 158. Let us also consider the formula $4+x^2$, which becomes a cube in two cases that may be considered as known; namely, $x=2$, and $x=11$. If now we first make $x=2+y$, the formula $8+4y+y^2$ will be required to become a cube, having for its root $2+\frac{1}{3}y$, and this cubed being $8+4y+\frac{2}{3}y^2+\frac{1}{27}y^3$, we find $1=\frac{2}{3}+\frac{1}{27}y$; therefore $y=9$, and $x=11$; which is the second given case. If we here suppose $x=11+y$, we shall have $4+x^2=125+22y+y^2$; which, being made equal to the cube of $5+py$, or to $125+75py+15p^2y^2+p^3y^3$, gives $p=\frac{22}{75}$; * See Article 247 of this Part.