Elements of Algebra

CHAP. IX. ADDITIONS. 587 x and y, it is evident, that, having $x + \alpha y = (x + \alpha y)^m$, we shall likewise have, $x + \beta y = (x + \beta y)^m$; wherefore, multiplying these two equations together, we shall have, $x^2 + axy + by^2 = (x^2 + axy + by^2)^m$; and, consequently, $z = x^2 + axy + by^2$. The problem, therefore, is solved. If $a$ were $= 0$, the foregoing formulae would become simpler; for we should have $A' = 1$, $A'' = 0$, $A''' = -b$, $A^{iv} = 0$, $A^v = b^2$, $A^{vi} = 0$, $A^{vii} = -b^3$, \ldots and, likewise, $B' = 0$, $B'' = b$, $B''' = 0$, $B^{iv} = -b^2$, $B^v = 0$, $B^{vi} = b^3$, \ldots Therefore, $x = x^m - \frac{m(m-1)}{2}x^{m-2}y^2b + \frac{m(m-1) \times (m-2) \times (m-3)}{2 \times 3 \times 4}x^{m-4}y^4b^2 -$, \ldots $y = mx^{m-1}y + \frac{m(m-1) \times (m-2)}{2 \times 3}x^{m-3}y^3b + \frac{m(m-1) \times (m-2) \times (m-3) \times (m-4)}{2 \times 3 \times 4 \times 5}x^{m-5}y^5b^2 +$, \ldots And these values will satisfy the equation, $x^2 + by^2 = (x^2 + by^2)^m$. 91. Let us now proceed to the formulae of three dimensions; in order to which, we shall denote by $\alpha$, $\beta$, $\gamma$, the three roots of the cubic equation, $s^3 - as^2 + bs - c = 0$, and we shall then consider the product of these three factors, $(x + \alpha y + \alpha^2 z) \times (x + \beta y + \beta^2 z) \times (x + \gamma y + \gamma^2 z)$, which must be rational, as we shall perceive. The multiplication being performed, we shall have the following product, $x^3 + (\alpha + \beta + \gamma)x^2y + (\alpha^2 + \beta^2 + \gamma^2)x^2z + (\alpha\beta + \alpha\gamma + \beta\gamma)xy^2 + (\alpha^2\beta + \alpha^2\gamma + \beta^2\alpha + \beta^2\gamma + \gamma^2\alpha + \gamma^2\beta)xyz + (\alpha^2\beta^2 + \alpha^2\gamma^2 + \beta^2\gamma^2)xz^2 + \alpha\beta\gamma y^3 + (\alpha^2\beta\gamma + \beta^2\alpha\gamma + \gamma^2\alpha\beta)y^2z + (\alpha^2\beta^2\gamma + \alpha^2\gamma^2\beta + \beta^2\gamma^2\alpha)yz^2 + \alpha^2\beta^2\gamma^2 z^3$. Now, from the nature of equations, we have $\alpha + \beta + \gamma = a$, $\alpha\beta + \alpha\gamma + \beta\gamma = b$, $\alpha\beta\gamma = c$. Farther, we shall find $\alpha^2 + \beta^2 + \gamma^2 = (\alpha + \beta + \gamma)^2 - 2(\alpha\beta + \alpha\gamma + \beta\gamma) = a^2 - 2b$, $\alpha^2\beta + \alpha^2\gamma + \beta^2\alpha + \beta^2\gamma + \gamma^2\alpha + \gamma^2\beta = (\alpha + \beta + \gamma) \times (\alpha\beta + \alpha\gamma + \beta\gamma) - 3\alpha\beta\gamma = ab - 3c$; and $\alpha^2\beta^2 + \alpha^2\gamma^2 + \beta^2\gamma^2 = (\alpha\beta + \alpha\gamma + \beta\gamma)^2 - 2(\alpha + \beta + \gamma)\alpha\beta\gamma = b^2 - 2ac$; also, $\alpha^2\beta\gamma + \beta^2\alpha\gamma + \gamma^2\alpha\beta = (\alpha + \beta + \gamma)\alpha\beta\gamma = ac$, and $\alpha^2\beta^2\gamma + \alpha^2\gamma^2\beta + \beta^2\gamma^2\alpha = (\alpha\beta + \alpha\gamma + \beta\gamma)\alpha\beta\gamma = bc$.