Elements of Algebra

264 ELEMENTS SECT. IV. such a manner, that $3\sqrt[3]{pq}$ may become equal to $f$, and $p+q=g$; for we then know that one of the roots of our equation will be $x = \sqrt[3]{p} + \sqrt[3]{q}$. 740. We have therefore to resolve these two equations, $3\sqrt[3]{pq} = f$, $p+q=g$. The first gives $\sqrt[3]{pq} = \frac{f}{3}$; or $pq = \frac{f^3}{27} = \frac{1}{27}f^3$, and $4pq = \frac{4}{27}f^3$. The second equation, being squared, gives $p^2 + 2pq + q^2 = g^2$; if we subtract from it $4pq = \frac{4}{27}f^3$, we have $p^2 - 2pq + q^2 = g^2 - \frac{4}{27}f^3$, and taking the square root of both sides, we have $p-q = \sqrt{g^2 - \frac{4}{27}f^3}$. Now, since $p+q=g$, we have, by adding $p+q$ to one side of the equation, and its equal, $g$, to the other, $2p = g + \sqrt{g^2 - \frac{4}{27}f^3}$; and, by subtracting $p-q$ from $p+q$, we have $2q = g - \sqrt{g^2 - \frac{4}{27}f^3}$; consequently, $p = \frac{g + \sqrt{g^2 - \frac{4}{27}f^3}}{2}$, and $q = \frac{g - \sqrt{g^2 - \frac{4}{27}f^3}}{2}$. 741. In a cubic equation, therefore, of the form $x^3 = fx+g$, whatever be the numbers $f$ and $g$, we have always for one of the roots $x = \sqrt[3]{\frac{g + \sqrt{g^2 - \frac{4}{27}f^3}}{2}} + \sqrt[3]{\frac{g - \sqrt{g^2 - \frac{4}{27}f^3}}{2}}$; that is, an irrational quantity, containing not only the sign of the square root, but also the sign of the cube root; and this is the formula which is called the Rule of Cardan. 742. Let us apply it to some examples, in order that its use may be better understood. Let $x^3 = 6x+9$. First, we shall have $f=6$, and $g=9$; so that $g^2=81$, $f^3=216$, $\frac{4}{27}f^3=32$; then $g^2 - \frac{4}{27}f^3 = 49$, and $\sqrt{g^2 - \frac{4}{27}f^3} = 7$. Therefore, one of the roots of the given equation is $x = \sqrt[3]{(\frac{9+7}{2})} + \sqrt[3]{(\frac{9-7}{2})} = \sqrt[3]{\frac{16}{2}} + \sqrt[3]{\frac{2}{2}} = \sqrt[3]{8} + \sqrt[3]{1} = \ldots$ $2+1=3$. 743. Let there be proposed the equation $x^3 = 3x+2$. Here, we shall have $f=3$ and $g=2$; and consequently, $g^2=4$, $f^3=27$, and $\frac{4}{27}f^3=4$; which gives $\sqrt{g^2 - \frac{4}{27}f^3} = 0$; whence it follows, that one of the roots is $x = \sqrt[3]{(\frac{2+0}{2})} + \sqrt[3]{(\frac{2-0}{2})} = 1+1=2$.