Elements of Algebra

CHAP. XV. OF ALGEBRA. 449 Therefore $r^4 = \frac{81}{16}s^4 + \frac{27}{2}s^3t + \frac{27}{2}s^2t^2 + 6st^3 + t^4$ $+ 4r^3s = \frac{27}{2}s^4 + 27s^3t + 18s^2t^2 + 4st^3$ $- 6r^2s^2 = -\frac{27}{2}s^4 - 18s^3t - 6s^2t^2$ $- 4rs^3 = -6s^4 - 4s^3t$ $+ s^4 = + s^4$; and, consequently, the formula will be $\frac{1}{16}s^4 + \frac{37}{2}s^3t + \frac{51}{2}s^2t^2 + 10st^3 + t^4$. This formula ought also to be a square, if multiplied by 16, by which means it becomes $s^4 + 296s^3t + 408s^2t^2 + 160st^3 + 16t^4$. Let us make this equal to the square of $s^2 + 148st - 4t^2$, that is, to $s^4 + 296s^3t + 21896s^2t^2 - 1184st^3 + 16t^4$; the first two terms, and the last, are destroyed on both sides, and we thus obtain the equation, $21896s - 1184t = 408s + 160t$, which gives $\frac{s}{t} = \frac{1344}{21488} = \frac{336}{5372} = \frac{84}{1343}$. Therefore, since $s=84$, and $t=1343$, we shall have $r = \frac{3}{2}s + t = 1469$, and, consequently, $x = r^4 - 6r^2s^2 + s^4 = 4565486027761$, and $y = 4r^3s - 4rs^3 = 1061652293520$. *** CHAP. XV. Solutions of some Questions, in which Cubes are required. 241. In the preceding chapter, we have considered some questions, in which it was required to transform certain formulæ into squares, and they afforded an opportunity of explaining several artifices requisite in the application of the rules which have been given. It now remains to consider questions, which relate to the transformation of certain formulæ into cubes; and the following solutions will throw some light on the rules, which have been already explained for transformations of this kind. 242. Question 1. It is required to find two cubes, $x^3$, and $y^3$, whose sum may be a cube. Since $x^3 + y^3$ must be a cube, if we divide this formula by $y^3$, the quotient ought likewise to be a cube, or $\frac{x^3}{y^3} + 1 = c$. If, therefore, $\frac{x}{y} = z - 1$, we shall have G G