Elements of Algebra

CHAP. XV. OF ALGEBRA. 461 $18^3 \times 36 + 18^3 \times 3y = 18^3 \times 25^2 + 25^3 y$, or $18^3 \times 36 - 18^3 \times 25^2 = 25^3 y - 18^3 \times 3y$; therefore $y = \frac{18^3 \times 36 - 18^3 \times 25^2}{25^3 - 3 \times 18^3} = \frac{18^3 \times (18 \times 36 - 25^2)}{25^3 - 3 \times 18^3}$; that is, $y = \frac{-324 \times 23}{1871} = \frac{-7452}{1871}$; and, consequently, $x = \frac{32}{1871}$. As it might be difficult to pursue this reduction in cubes, it is proper to observe, that the question may always be reduced to squares. In fact, since $3x(x^2 + 2)$ must be a cube, let us suppose $3x(x^2 + 2) = x^3 y^3$; dividing by $x$, we shall have $3x^2 + 6 = x^2 y^3$; and, consequently, $x^2 = \frac{6}{y^3-3} = \frac{36}{6y^3-18}$. Now, the numerator of this fraction being already a square, it is only necessary to transform the denominator, $6y^3 - 18$, into a square, which also requires that we have already found a case. For this purpose, let us consider that 18 is divisible by 9, but 6 only by 3, and that $y$ therefore may be divided by 3; if we make $y = 3z$, our denominator will become $162z^3 - 18$, which being divided by 9, and becoming $18z^3 - 2$, must still be a square. Now, this is evidently true of the case $z=1$. So that we shall make $z = 1+v$, and we must have $16 + 54v + 54v^2 + 18v^3 = \square$. Let its root be $4 + \frac{27}{4}v$, the square of which is $16 + 54v + \frac{729}{16}v^2$, and we must have $54 + 18v = \frac{729}{16}$; or $18v = -\frac{135}{16}$, or $2v = -\frac{15}{16}$; and, consequently, $v = -\frac{15}{32}$; which produces $z = 1+v = \frac{17}{32}$, and then $y = \frac{51}{32}$. Let us now resume the denominator $6y^3 - 18 = 162z^3 - 18 = 9(18z^3 - 2)$; and since the square root of the factor, $18z^3 - 2$, is $4 + \frac{27}{4}v = \frac{107}{128}$, that of the whole denominator is $\frac{321}{128}$: but the root of the numerator is 6; therefore $x = \frac{6}{\frac{321}{128}} = \frac{256}{107}$, a value quite different from that which we found before. It follows, therefore, that the roots of our three cubes sought are $x-1 = \frac{149}{107}$, $x = \frac{256}{107}$, $x+1 = \frac{363}{107}$: and the sum of the cubes of these three numbers will be a cube, whose root, $xy, = \frac{256}{107} \times \frac{51}{32} = \frac{13056}{3424} = \frac{408}{107}$. 250. We shall here finish this Treatise on the Indeterminate Analysis, having had sufficient occasion, in the questions which we have resolved, to explain the chief artifices that have hitherto been devised in this branch of Algebra.