Elements of Algebra

266 ELEMENTS SECT. IV. tion of which it is evident, since we immediately perceive that it is the product of the factors $y(y^2 - 1) = y (y + 1) \times (y - 1) = 0$. If we now make each of these factors $= 0$, we have 1 $\begin{cases} y = 0, \\ x = 2, \end{cases}$ 2 $\begin{cases} y = -1, \\ x = 1, \end{cases}$ 3 $\begin{cases} y = 1, \\ x = 3, \end{cases}$ that is to say, the three roots which we have already found. 746. Let there now be given the general equation of the third degree, $x^3 + ax^2 + bx + c = 0$, of which it is required to destroy the second term. For this purpose, we must add to $x$ the third of the co-efficient of the second term, preserving the same sign, and then write for this sum a new letter, as for example $y$, so that we shall have $x + \frac{1}{3}a = y$, and $x = y - \frac{1}{3}a$; whence results the following calculation: $x = y - \frac{1}{3}a$, $x^2 = y^2 - \frac{2}{3}ay + \frac{1}{9}a^2$, and $x^3 = y^3 - ay^2 + \frac{1}{3}a^2y - \frac{1}{27}a^3$; Consequently, $x^3 = y^3 - ay^2 + \frac{1}{3}a^2y - \frac{1}{27}a^3$ $ax^2 = ay^2 - \frac{2}{3}a^2y + \frac{1}{9}a^3$ $bx = by - \frac{1}{3}ab$ $c = c$ or, $y^3 - (\frac{1}{3}a^2 - b)y + \frac{2}{27}a^3 - \frac{1}{3}ab + c = 0$, an equation in which the second term is wanting. 747. We are enabled, by means of this transformation, to find the roots of all equations of the third degree, as the following example will shew. Let it be proposed to resolve the equation $x^3 - 6x^2 + 13x - 12 = 0$. Here it is first necessary to destroy the second term; for which purpose, let us make $x - 2 = y$, and then we shall have $x = y + 2$, $x^2 = y^2 + 4y + 4$, and $x^3 = y^3 + 6y^2 + 12y + 8$; therefore, $x^3 = y^3 + 6y^2 + 12y + 8$ $-6x^2 = -6y^2 - 24y - 24$ $13x = 13y + 26$ $-12 = -12$ which gives $y^3 + y - 2 = 0$; or $y^3 = -y + 2$. And if we compare this equation with the formula, (Art. 741) $x^3 = fx + g$, we have $f = -1$, and $g = 2$; wherefore, $g^2 = 4$, and $\frac{4}{27}f^3 = -\frac{4}{27}$; also, $g^2 - \frac{4}{27}f^3 = 4 + \frac{4}{27} = \frac{112}{27}$; and $\sqrt{(g^2 - \frac{4}{27}f^3)} = \sqrt{\frac{112}{27}} = \frac{4\sqrt{21}}{9}$; consequently,