Elements of Algebra

CHAP. XII. OF ALGEBRA. 271 5. Given $y^3 + 3y^2 + 9y = 13$, to determine $y$. Ans. $y = 1$. 6. Given $x^3 - 6x = -9$, to find the value of $x$. Ans. $x = -3$. 7. Given $x^3 - 6x^2 + 10x = 8$, to find $x$. Ans. $x = 4$. 8. Given $p^3 - \frac{19}{3}p = \frac{150}{27}$, to find $p$. Ans. $p = 8\frac{1}{3}$. 9. Given $x^3 - \frac{7}{3}x = \frac{70}{27}$, to find $x$. Ans. $x = 2\frac{1}{3}$. 10. Given $a^3 - 91a = -330$, to find $a$. Ans. $a = 5$. 11. Given $y^3 - 19y = 30$, what is the value of $y$? Ans. $y = 5$. Taking the value of $x$, answering to log. $x$, negatively in the first equation, and positively in the two latter. As an example of this mode of solution, in what is usually called the Irreducible Case of Cubic Equations, Let $x^3 - 3x = 1$, to find its 3 roots. Here $\frac{q}{2} (\frac{3}{p})^{\frac{3}{2}} = \frac{1}{2} (\frac{3}{3})^{\frac{3}{2}} = \frac{1}{2} = .5 = \cos. 60^\circ = z$, hence $x = 2\sqrt{\frac{p}{3}} \times \cos. \frac{z}{3} = 2 \cos. 20^\circ = 1.8793852$ $x = -2\sqrt{\frac{p}{3}} \times \cos.(60^\circ - \frac{z}{3}) = -2 \cos. 40^\circ = -1.5320888$ $x = -2\sqrt{\frac{p}{3}} \times \cos.(60^\circ + \frac{z}{3}) = -2 \cos. 80^\circ = -0.3472964$. Also, let $x^3 - 3x = -1$, to find its three roots. Here, as before, $\frac{q}{2} (\frac{3}{p})^{\frac{3}{2}} = .5 = \cos. 60^\circ = z$, hence $x = -2\sqrt{\frac{p}{3}} \times \cos. \frac{z}{3} = -2 \cos. 20^\circ = -1.8793852$ $x = -2\sqrt{\frac{p}{3}} \times \cos. (60^\circ - \frac{z}{3}) = 2 \cos. 40^\circ = 1.5320888$ $x = -2\sqrt{\frac{p}{3}} \times \cos. (60^\circ + \frac{z}{3}) = 2 \cos. 80^\circ = 0.3472964$. Where the roots are the negatives of those of the first case. For the mode of investigating these kinds of formulæ, see, in addition to the references already given, Cagnoli, Traité de Trigon. and Article Irreducible Case, in the Supplement to Dr. Hutton's Mathematical Dictionary.