Elements of Algebra

386 ELEMENTS PART II. having previously found a satisfactory case; so that supposing $q = pz$, we must have the formula, $2p^2z(7p^2 + p^2z^2) = 2p^4z(7+z^2)$, and, consequently, if we divide by $p^4$, the formula $2z(7+z^2)$ must become a square. The known case is here $z=1$, for which reason we shall make $z=1+y$, and we shall thus have $(2+2y) \times (8+2y+y^2) = 16+20y+6y^2+2y^3$, the root of which we shall suppose to be $4+\frac{5}{2}y$; then its square will be $16+20y+\frac{25}{4}y^2$, which, being made equal to the formula, gives $6+2y=\frac{25}{4}$; therefore $y=\frac{1}{8}$, and $z=\frac{9}{8}$. Also, $z=\frac{q}{p}$; so that $q=9$, and $p=8$, which makes $x = \frac{36}{7}$, and the formula $7+x^2 = \frac{279841}{20736}$. If we now extract the square root of this fraction, we find $\frac{529}{144}$; and taking the square root of this also, we find $\frac{23}{12}$; consequently, the given formula is the biquadrate of $\frac{2}{3}$. 161. Before we conclude this chapter, we must observe, that there are some formulæ, which may be transformed into cubes in a general manner; for example, if $cx^2$ must be a cube, we have only to make its root $=px$, and we find $cx^2=p^3x^3$, or $c=p^3x$, that is, $x=\frac{c}{p^3}$, or $x=cq^3$, if we write $\frac{1}{q}$ instead of $p$. The reason of this evidently is, that the formula contains a square, on which account, all such formulæ, as $a(b+cx)^2$, or $ab^2+2abcx+ac^2x^2$, may very easily be transformed into cubes. In fact, if we suppose its cube root to be $\frac{b+cx}{q}$, we shall have the equation $a(b+cx)^2 = \frac{(b+cx)^3}{q^3}$, which, divided by $(b+cx)^2$, gives $a = \frac{b+cx}{q^3}$, whence we get $x = \frac{aq^3-b}{c}$, a value in which $q$ is arbitrary. This shews how useful it is to resolve the given formulæ into their factors, whenever it is possible: on this subject, therefore, we think it will be proper to dwell at some length in the following chapter.