Elements of Algebra

CHAP. IX. ADDITIONS. 589 Therefore, by multiplying these three equations together, we shall have, on the one side, the product of the two given formulæ, and on the other, the formula, $x'^3 + ax'^2y' + (a^2 - 2b)x'^2z' + bx'y'^2 + (ab - 3c)x'y'z' + (b^2 - 2ac)x'z'^2 + cy'^3 + acy'^2z' + bcy'z'^2 + c^2z'^3$, which will therefore be equal to the product required; and is evidently of the same form as each of the two formulæ of which it is composed. If we had a third formula, such as $x''^3 + ax''^2y'' + (a-2b)x''^2z'' + bx''y''^2 + (ab - 3c)x''y''z'' + (b^2 - 2ac)x''z''^2 + cy''^3 + acy''z'' + bcy''z''^2 + c^2z''^3$ and if we wished to have the product of this formula and the two preceding, it is evident, that we should only have to make $x' = xx'' - c(yz'' + zy'') + aczz''$, $y' = xy'' + yx'' - b(yz'' + zy'') - (ab - c)zz''$, $z' = xz'' + zx'' + yy'' + a(yz'' + zy'') + (a^2 - b)zz''$, and we should have, for the product required, $x'^3 + ax'^2y' + (a^2 - 2b)x'^2z' + bx'y'^2 + (ab - 3c)x'y'z' + (b^2 - 2ac)x'z'^2 + cy'^3 + acy'^2z' + bcy'z'^2 + c^2z'^3$. 92. Let us now make $x' = x$, $y' = y$, $z' = z$, and we shall have, $X = x^2 - 2cyz + acz^2$, $Y = 2xy - 2byz - (ab - c)z^2$, $Z = 2xz + y^2 + 2ayz + (a^2 - b)z^2$; and these values will satisfy the equation, $X^3 + aX^2Y + bXY^2 + cY^3 + (a^2 - 2b)X^2Z + (ab - 3c)XYZ + acY^2Z + (b^2 - 2ac)XZ^2 + bcYZ^2 + c^2Z^3 = v^3$, by taking $v = x^3 + ax^2y + bxy^2 + cy^3 + (a^2 - 2b)x^2z + (ab - 3c)xyz + acy^2z + (b^2 - 2ac)xz^2 + bcyz^2 + c^2z^3$. Wherefore, if we had, for example, to resolve an equation of this form, $x^3 + ax^2y + bxy^2 + cy^3 = v^3$, a, b, c, being any given quantities, we should only have to destroy z, by making $2xz + y^2 + 2ayz + (a^2 - b^2)z^2 = 0$, whence we derive $x = - \frac{y^2 + 2ayz + (a^2 - b^2)z^2}{2z}$; and, substituting this