Elements of Algebra

CHAP. XII. OF ALGEBRA. 269 QUESTIONS FOR PRACTICE. 1. Given $y^3 + 30y = 117$, to determine $y$. Ans. $y=3$. 2. Given $y^3 - 36y = 91$, to find the value of $y$. Ans. $y=7$. $\frac{1}{2}(\log. \tan. (45^o - \frac{1}{2}z) + 20) = \log. \tan. u$, Then $\log. x = \frac{1}{3}\log. \frac{4p}{3} + \log. \cot. 2u - 10$. 2. $x^3 + px + q = 0$. Put $\frac{q}{2}(\frac{3}{p})^{\frac{3}{2}} = \tan. z$, and $\sqrt[2]{(\tan. (45^o - \frac{1}{2}z))} = \tan. u$, Then $x = -2\sqrt{\frac{p}{3}} \times \cot. 2u$. Or, putting Log. $\frac{q}{2} + 10 - \frac{1}{3}\log. \frac{p}{3} = \log. \tan. z$, and $\frac{1}{2}(\log. \tan. (45^o - \frac{1}{2}z) + 20) = \log. \tan. u$, Then $\log. x = 10 - \frac{1}{3}\log. \frac{4p}{3} - \log. \cot. 2u$. 3. $x^3 - px - q = 0$. This form has 2 cases, according as $\frac{2}{q}(\frac{p}{3})^{\frac{3}{2}}$ is less, or greater than 1. In the 1st case, put $\frac{2}{q}(\frac{p}{3})^{\frac{3}{2}} = \cos. z$. And $\sqrt[2]{(\tan. (45^o - \frac{1}{2}z))} = \tan. u$; Then $x = 2\sqrt{\frac{p}{3}} \times \operatorname{cosec}. 2u$. Or, putting $10 + \frac{1}{3}\log. \frac{p}{3} - \log. \frac{q}{2} = \log. \cos. z$, and $\frac{1}{2}(\log. \tan. (45^o - \frac{1}{2}z) + 20) = \log. \tan. u$; Then $\log. x = 10 + \log. \frac{4p}{3} - \log. \sin. 2u$. In the 2d case, put $\frac{q}{2}(\frac{3}{p})^{\frac{3}{2}} = \cos. z$, and $x$ will have the 3 following values: $x = +2\sqrt{\frac{p}{3}} \times \cos. \frac{z}{3}$ $x = -2\sqrt{\frac{p}{3}} \times \cos. (60^o - \frac{z}{3})$ $x = -2\sqrt{\frac{p}{3}} \times \cos. (60^o + \frac{z}{3})$ or,