Elements of Algebra

586 ADDITIONS. CHAP. IX. this quantity may become a power of the dimension $m$, each of its factors must likewise become a similar power. Let us, therefore, first make $x + \alpha y = (x + \alpha y)^m$, and, expressing this power by Newton's theorem, we shall have $x^m + mx^{m-1}y\alpha + \frac{m(m-1)}{2} x^{m-2}y^2\alpha^2 + \frac{m(m-1) \times (m-2)}{2 \times 3} x^{m-3}y^3\alpha^3 + \ldots$ Now, since $\alpha$ is one of the roots of the equation, $\alpha^2 - a\alpha + b = 0$, we shall also have $\alpha^2 - a\alpha + b = 0$; wherefore, $\alpha^2 = a\alpha - b$, $\alpha^3 = a\alpha^2 - b\alpha = (a^2 - b)\alpha - ab$, $\alpha^4 = (a^2 - b)\alpha^2 - ab\alpha = (a^3 - 2ab)\alpha - a^2b + b^2$; and so on. Thus, we shall only have to substitute these values in the preceding formula, and then we shall find it to be compounded of two parts, the one wholly rational, which we shall compare to $x$, and the other wholly multiplied by the root $\alpha$, which we shall compare to $\alpha y$. If, in order to simplify, we make $A' = 1$ $B' = 0$ $A'' = a$ $B'' = b$ $A''' = aA'' - bA'$ $B''' = aB'' - bB'$ $A^{IV} = aA''' - bA''$ $B^{IV} = aB''' - bB''$ $A^V = aA^{IV} - bA'''$, $B^V = aB^{IV} - bB'''$, $\ldots, \ldots, \ldots$ we shall have, $\alpha = A'\alpha - B'$ $\alpha^2 = A''\alpha - B''$ $\alpha^3 = A'''\alpha - B'''$ $\alpha^4 = A^{IV}\alpha - B^{IV}, \ldots$ Wherefore, substituting these values, and comparing, we shall have $x = x^m - mx^{m-1}yB' - \frac{m(m-1)}{2}x^{m-2}y^2B'' - \frac{m(m-1) \times (m-2)}{2 \times 3}x^{m-3}y^3B''' - \ldots$ $Y = mx^{m-1}yA' + \frac{m(m-1)}{2}x^{m-2}y^2A'' + \frac{m(m-1) \times (m-2)}{2 \times 3}x^{m-3}y^3A''' + \ldots$ Now, as the root $\alpha$ does not enter into the expressions of