Elements of Algebra

$2u(t^2 - u^2) = 2u(t + u) \times (t - u)$ a cube; and as these three factors are prime to each other, each must of itself be a cube. Suppose therefore $t + u = f^3$, and $t - u = g^3$, we shall have $2u = f^3 - g^3$; that is to say, if $2u$ were a cube, $f^3 - g^3$ would be a cube. We should consequently have two cubes, $f^3$ and $g^3$, much smaller than the first, whose difference would be a cube, and that would enable us also to find two cubes whose sum would be a cube; since we should only have to make $f^3 - g^3 = h^3$, in order to have $f^3 = h^3 + g^3$, or a cube equal to the sum of two cubes. Thus, the foregoing conclusion is fully confirmed; for as we cannot assign, in great numbers, two cubes whose sum or difference is a cube, it follows from what has been before observed, that no such cubes are to be found among small numbers. 244. Since it is impossible, therefore, to find two cubes, whose sum or difference is a cube, our first question falls to the ground: and, indeed, it is more usual to enter on this subject with the question of determining three cubes, whose sum may make a cube; supposing, however, two of those cubes to be arbitrary, so that it is only required to find the third. We shall therefore proceed immediately to this question. 245. Question 2. Two cubes $a^3$, and $b^3$, being given, required a third cube, such, that the three cubes added together may make a cube. It is here required to transform into a cube the formula $a^3 + b^3 + x^3$; which cannot be done unless we already know a satisfactory case; but such a case occurs immediately; namely, that of $x = -a$. If therefore we make $x = y - a$, we shall have $x^3 = y^3 - 3ay^2 + 3a^2y - a^3$; and, consequently, it is the formula $y^3 - 3ay^2 + 3a^2y + b^3$ that must become a cube. Now, the first and the last term here being cubes, we immediately find two solutions. 1. The first requires us to represent the root of the formula by $y + b$, the cube of which is $y^3 + 3by^2 + 3b^2y + b^3$; and we thus obtain $-3ay^2 + 3a^2y = 3by^2 + 3b^2y$; and, consequently, $y = \frac{a^2 - b^2}{a + b} = a - b$; but $x = -b$, so that this solution is of no use. 2. But we may also represent the root by $fy + b$, the cube of which is $f^3y^3 + 3bf^2y^2 + 3b^2fy + b^3$, and then determine $f$ in such a manner, that the third terms may be destroyed, namely, by making $3a^2 = 3b^2f$, or $f = \frac{a^2}{b^2}$; for