Elements of Algebra

CHAP. XII. OF ALGEBRA. 267 $y = (\frac{2+\frac{4\sqrt{21}}{9}}{2})^{\frac{1}{3}} + (\frac{2-\frac{4\sqrt{21}}{9}}{2})^{\frac{1}{3}}$, or $y = \sqrt[3]{(1 + \frac{2\sqrt{21}}{9})} + \sqrt[3]{(1 - \frac{2\sqrt{21}}{9})}$, or $y = \sqrt[3]{(9 + \frac{2\sqrt{21}}{9})} + \sqrt[3]{(9 - \frac{2\sqrt{21}}{9})}$ $y = \sqrt[3]{(\frac{27+6\sqrt{21}}{27})} + \sqrt[3]{(\frac{27-6\sqrt{21}}{27})}$ or $y = \frac{1}{3}\sqrt[3]{(27+6\sqrt{21})} + \frac{1}{3}\sqrt[3]{(27-6\sqrt{21})}$; and it remains to substitute this value in $x = y + 2$. 748. In the solution of this example, we have been brought to a quantity doubly irrational; but we must not immediately conclude that the root is irrational: because the binomials $27 \pm 6\sqrt{21}$ might happen to be real cubes; and this is the case here; for the cube of $\frac{3+\sqrt{21}}{2}$ being $\frac{216+48\sqrt{21}}{8} = 27+6\sqrt{21}$, it follows that the cube root of $27+6\sqrt{21}$ is $\frac{3+\sqrt{21}}{2}$, and that the cube root of $27-6\sqrt{21}$ is $\frac{3-\sqrt{21}}{2}$. Hence the value which we found for $y$ becomes $y = \frac{1}{3}(\frac{3+\sqrt{21}}{2}) + \frac{1}{3}(\frac{3-\sqrt{21}}{2}) = \frac{1}{2} + \frac{1}{2} = 1$. Now, since $y=1$, we have $x=3$ for one of the roots of the equation proposed, and the other two will be found by dividing the equation by $x-3$. $x - 3) x^3 - 6x^2 + 13x - 12 (x^2 - 3x + 4$ $\underline{x^3 - 3x^2 \quad\quad\quad}$ $\quad -3x^2 + 13x$ $\quad \underline{-3x^2 + 9x \quad}$ $\quad \quad \quad 4x - 12$ $\quad \quad \quad \underline{4x - 12}$ $\quad \quad \quad \quad 0.$ Also making the quotient $x^2 - 3x + 4 = 0$, we have $x^2 = 3x - 4$; and