Elements of Algebra

590 ADDITIONS. CHAP. IX. value of $x$ in the foregoing expressions of $x$, $y$, and $v$, we shall have very general values of these quantities, which will satisfy the equation proposed. This solution deserves particular attention, on account of its generality, and the manner in which we have arrived at it; which is, perhaps, the only way in which it can be easily resolved. We should likewise obtain the solution of the equation, $x^3 + ax^2y' + (a^2 - 2b)x^2z' + bxy'^2 + (ab - 3c)xy'z' + (b^2 - 2ac)xz'^2 + cy'^3 + acy'^2z' + bcy'z'^2 + c^2z'^3 = v'^3$, by making, in the foregoing formulæ, $x'' = x' = x$, $y'' = y' = y$, $z'' = z' = z$, and taking $v = x^3 + ax^2y + (a^2 - 2b)x^2z + bxy^2 + (ab - 3c)xyz + (b^2 - 2ac)xz^2 + cy^3 + acy^2z + bcyz^2 + c^2z^3$. And we might resolve, successively, the cases in which, instead of the third power $v^3$, we should have $v^4$, $v^5$, \ldots. But we are going to consider these questions in a general manner, as we have done Art. 90. 93. Let it be proposed, therefore, to resolve an equation of this form, $x^3 + ax^2y + (a^2 - 2b)x^2z + bxy^2 + (ab - 3c)xyz + (b^2 - 2ac)xz^2 + cy^3 + acy^2z + bcyz^2 + c^2z^3 = v^m$. Since the quantity, which forms the first side of this equation, is nothing more than the product of these three factors, $(x + \alpha y + \alpha^2 z) \times (x + \beta y + \beta^2 z) \times (x + \gamma y + \gamma^2 z)$, it is evident that, in order to render this quantity equal to a power of the dimension $m$, we have only to make each of its factors separately equal to such a power. Let then $x + \alpha y + \alpha^2 z = (x + ay + a^2 z)^m$. We shall begin by expressing the mth power of $x + ay + a^2z$ according to Newton's theorem, which will give $x^m + mx^{m-1}(y + az)a + \frac{m(m-1)}{2}x^{m-2}(y + az)^2a^2 + \frac{m(m-1) \times (m-2)}{2 \times 3}x^{m-3}(y + az)^3a^3 + \ldots$. Or rather, forming the different powers of $y + az$, and then arranging them, according to the dimensions of $a$,