Elements of Algebra

CHAP. XV. OF ALGEBRA. 455 we thus arrive at the equation $y - 3a = \frac{a^6y}{b^6} + \frac{3a^4}{b^3}$, which multiplied by $b^6$, becomes $b^6y - 3ab^6 = a^6y + 3a^4b^3$. This gives $y = \frac{3a^4b^3+3ab^6}{b^6-a^6} = \frac{3ab^3(a^3+b^3)}{b^6-a^6} = \frac{3ab^3}{b^3-a^3}$; and, consequently, $x = y - a = \frac{2ab^3+a^4}{b^3-a^3} = a \times \frac{2b^3+a^3}{b^3-a^3}$. So that the two cubes $a^3$ and $b^3$ being given, we know also the root of the third cube sought; and if we would have that root positive, we have only to suppose $b^3$ to be greater than $a^3$. Let us apply this to some examples. 1. Let 1 and 8 be the two given cubes, so that $a=1$, and $b=2$; the formula $9+x^3$ will become a cube, if $x = \frac{17}{7}$; for we shall have $9+x^3 = \frac{8000}{343} = (\frac{20}{7})^3$. 2. Let the given cubes be 8 and 27, so that $a=2$, and $b=3$; the formula $35+x^3$ will be a cube, when $x = \frac{124}{19}$. 3. If 27 and 64 be the given cubes, that is, if $a=3$, and $b=4$, the formula $91+x^3$ will become a cube, if $x = \frac{465}{37}$. And, generally, in order to determine third cubes for any two given cubes, we must proceed by substituting $\frac{2ab^3+a^4}{b^3-a^3} + z$ instead of $x$, in the formula $a^3+b^3+x^3$; for by these means we shall arrive at a formula like the preceding, which would then furnish new values of $z$; but it is evident that this would lead to very prolix calculations. 246. In this question, there likewise occurs a remarkable case; namely, that in which the two given cubes are equal, or $a=b$; for then we have $x = \frac{3a^4}{0} = \infty$; that is, we have no solution; and this is the reason why we are not able to resolve the problem of transforming into a cube the formula $2a^3+x^3$. For example, let $a=1$, or let this formula be $2+x^3$, we shall find that whatever forms we give it, it will always be to no purpose, and we shall seek in vain for a satisfactory value of $x$. Hence, we may conclude with sufficient certainty, that it is impossible to find a cube equal to the sum of a cube, and of a double cube; or that the equation $2a^3+x^3=y^3$ is impossible. As this equation