Elements of Algebra

CHAP. X. OF ALGEBRA. 385 But we may also suppose $x = \frac{2+11y}{1-y}$; and, in this case, our formula becomes $4 + \frac{4+44y+121y^2}{1-2y+y^2} = \frac{8+36y+125y^2}{(1-y)^3}$; so that multiplying both terms by $1-y$, we have $8+28y+89y^2-125y^3$ to transform into a cube. If we therefore suppose, according to the first case, the root to be $2+\frac{7}{3}y$, the cube of which is $8+28y+\frac{98}{3}y^2+\frac{343}{27}y^3$, we have $89-125y = \frac{98}{3}+\frac{343}{27}y$, or $\frac{1718}{27}y = \frac{169}{3}$; and, consequently, $y = \frac{1521}{3718} = \frac{9}{22}$; whence we get $x=11$; that is, one of the values already known. But let us rather consider our formula with reference to the third case, and suppose its root to be $2-5y$; the cube of this binomial being $8-60y+150y^2-125y^3$, we shall have $28+89y = -60+150y$; therefore $y=\frac{88}{61}$, whence we get $x=-\frac{1090}{27}$; so that our formula becomes $\frac{1191016}{729}$, or the cube of $\frac{106}{9}$. 160. The foregoing are the methods at present known for reducing such formulæ as we have considered, either to squares, or to cubes, provided the highest power of the unknown quantity do not exceed the fourth power in the former case, nor the third in the latter. We might also add the problem for transforming a given formula into a biquadrate, in the case of the unknown quantity not exceeding the second degree. But it will be perceived, that, if such a formula as $a+bx+cx^2$ were proposed to be transformed into a biquadrate, it must in the first place be a square; after which it will only remain to transform the root of that square into a new square, by the rules already given. If $x^2+7$, for example, is to be made a biquadrate, we first make it a square, by supposing $x = \frac{7p^2-q^2}{2pq}$, or $x = \frac{q^2-7p^2}{2pq}$; the formula then becomes equal to the square, $\frac{q^4-14q^2p^2+49p^4}{4p^2q^2} + 7 = \frac{q^4+14q^2p^2+49p^4}{4p^2q^2}$, the root of which, $\frac{7p^2+q^2}{2pq}$, must likewise be transformed into a square; for this purpose, let us multiply the two terms by $2pq$, in order that the denominator becoming a square, we may have only to consider the numerator $2pq(7p^2+q^2)$. Now, we cannot make a square of this formula, without c c