Elements of Algebra

588 ADDITIONS. CHAP. IX. Therefore, making these substitutions, the product in question will be $x^3+ax^2y + (a^2-2b)x^2z+bxy^2+(ab-3c)xyz+(b^2-2ac)xz^2+cy^3+acy^2z + bcyz^2+c^2z^3$. And this formula will have the property, that if we multiply together as many similar formulæ as we choose, the product will always be a similar formula. Let us suppose that the product of the foregoing formula by the following was required, namely, $x'^3 + ax'^2y' + (a^2-2b)x'^2z' + bx'y'^2 + (ab-3c)x'y'z' + (b^2-2ac)x'z'^2 + cy'^3 + acy'^2z' + bcy'z'^2 + c^2z'^3$; it is evident, that we have only to seek the product of these six factors, $x + \alpha y + \alpha^2 z$, $x + \beta y + \beta^2 z$, $x + \gamma y + \gamma^2 z$, $x' + \alpha y' + \alpha^2 z'$, $x' + \beta y' + \beta^2 z'$, $x' + \gamma y' + \gamma^2 z'$; if we first multiply $x + \alpha y + \alpha^2 z$, by $x' + \alpha y' + \alpha^2 z'$, we shall have this partial product, $xx'+\alpha(xy'+yx')+\alpha^2(xz'+zx'+yy')+\alpha^3(yz'+zy')+\alpha^4zz'$. Now, $\alpha$ being one of the roots of the equation, $s^3 - as^2 + bs - c = 0$, we shall have $\alpha^3 - a\alpha^2 + b\alpha - c = 0$; consequently, $\alpha^3 = a\alpha^2 - b\alpha + c$; whence, $\alpha^4 = a\alpha^3 - b\alpha^2 + c\alpha = (a^2-b)\alpha^2 - (ab-c)\alpha + ac$; so that substituting these values, and, in order to abridge, making $x = xx' - c(yz' + zy') + aczz'$, $y = xy' + yx' - b(yz' + zy') - (ab-c)zz'$, $z = xz' + zx' + yy' + a(yz' + zy') + (a^2-b)zz'$, the product in question will become of this form, $x + \alpha y + \alpha^2 z$; that is to say, of the same form as each of those from which it has been produced. Now, as the root $\alpha$ does not enter into the values of x, y, z, it is evident, that these quantities will be the same, if we change $\alpha$ into $\beta$, or $\gamma$; wherefore, since we already have $(x + \alpha y + \alpha^2 z) \times (x' + \alpha y' + \alpha^2 z') = x + \alpha y + \alpha^2 z$, we shall likewise have, by changing $\alpha$ into $\beta$, $(x + \beta y + \beta^2 z) \times (x' + \beta y' + \beta^2 z') = x + \beta y + \beta^2 z$; and, by changing $\alpha$ into $\gamma$, $(x + \gamma y + \gamma^2 z) \times (x' + \gamma y' + \gamma^2 z') = x + \gamma y + \gamma^2 z$.