Elements of Algebra

CHAP. IX. ADDITIONS. 591 $x^m + mx^{m-1}ya + (mx^{m-1}z + \frac{m(m-1)}{2}x^{m-2}y^2)a^2 + (m(m-1)x^{m-2}yz + \frac{m(m-1) \times (m-2)}{2 \times 3}x^{m-3}y^3)a^3 + \ldots$ But as in this formula we do not easily perceive the law of the terms, we shall suppose, in general, $(x + ay + a^2z)^m = P + P'a + P''a^2 + P'''a^3 + P^{iv}a^4 + \ldots$ and we shall find, $P = x^m$, $P' = \frac{myP}{x}$, $P'' = \frac{(m-1)yP' + 2mzP}{2x}$, $P''' = \frac{(m-2)yP'' + (2m-1)zP'}{3x}$, $P^{iv} = \frac{(m-3)yP''' + (2m-2)zP''}{4x}, \ldots$ which may easily be demonstrated by the differential calculus. Now, since $\alpha$ is one of the roots of the equation, $s^3 - as^2 + bs - c = 0$, we shall have $\alpha^3 - a\alpha^2 + b\alpha - c = 0$; whence, $\alpha^3 = a\alpha^2 - b\alpha + c$; wherefore, $\alpha^4 = a\alpha^3 - b\alpha^2 + c\alpha = (a^2 - b)\alpha^2 - (ab - c)\alpha + ac$, $\alpha^5 = (a^2 - b)\alpha^3 - (ab - c)\alpha^2 + ac\alpha = (a^3 - 2ab + c)\alpha^2 - (a^2b - b^2 - ac)z + (a^2 - b)c$; and so on. So that if, in order to simplify, we make $A' = 0 \qquad A^{iv} = aA''' - bA'' + cA' \qquad c' = 0$ $A'' = 1 \qquad A^v = aA^{iv} - bA''' + cA'' \qquad c'' = 0$ $A''' = a \qquad A^{vi} = aA^v - bA^{iv} + cA''', \ldots \qquad c''' = c$ $B' = 1 \qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad c^{iv} = ac''' - bc'' + cc'$ $B'' = 0 \qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad c^v = ac^{iv} - bc''' + cc''$ $B''' = b \qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad c^{vi} = ac^v - bc^{iv} + cc''', \ldots$ $B^{iv} = aB''' - bB'' + cB'$ $B^v = aB^{iv} - bB''' + cB''$ $B^{vi} = aB^v - bB^{iv} + cB''', \ldots$ we shall have, $\alpha = A'\alpha^2 - B'\alpha + c' \qquad \alpha^3 = A'''\alpha^2 - B'''\alpha + c'''$ $\alpha^2 = A''\alpha^2 - B''\alpha + c'' \qquad \alpha^4 = A^{iv}\alpha^2 - B^{iv}\alpha + c^{iv}, \ldots$ Substituting these values, therefore, in the expression