Elements of Algebra

CHAP. X. OF ALGEBRA. 249 quantity, and when neither the square of the unknown quantity, nor the unknown quantity itself, is found in the equation; so that $x^3 = 125$; or, more generally, $x^3 = a$, $x^3 = \frac{a}{b}$, &c. are equations of this kind. 707. It is evident how we are to deduce the value of $x$ from such an equation, since we have only to extract the cube root of both sides. Thus, the equation $x^3 = 125$ gives $x = 5$, the equation $x^3 = a$ gives $x = \sqrt[3]{a}$, and the equation $x^3 = \frac{a}{b}$ gives $x = \sqrt[3]{\frac{a}{b}}$, or $x = \frac{\sqrt[3]{a}}{\sqrt[3]{b}}$. To be able, therefore, to resolve such equations, it is sufficient that we know how to extract the cube root of a given number. 708. But in this manner, we obtain only one value for $x$: but since every equation of the second degree has two values, there is reason to suppose that an equation of the third degree has also more than one value. It will be deserving our attention to investigate this; and, if we find that in such equations $x$ must have several values, it will be necessary to determine those values. 709. Let us consider, for example, the equation $x^3 = 8$, with a view of deducing from it all the numbers, whose cubes are, respectively, 8. As $x = 2$ is undoubtedly such a number, what has been said in the last chapter shews that the quantity $x^3 - 8 = 0$, must be divisible by $x - 2$: let us therefore perform this division. $x - 2) x^3 - 8 (x^2 + 2x + 4$ $x^3 - 2x^2$ ————— $2x^2 - 8$ $2x^2 - 4x$ ————— $4x - 8$ $4x - 8$ ————— $0.$ Hence it follows, that our equation, $x^3 - 8 = 0$, may be represented by these factors; $(x - 2) \times (x^2 + 2x + 4) = 0$. 710. Now, the question is, to know what number we are to substitute instead of $x$, in order that $x^3 = 8$, or that $x^3 - 8 = 0$; and it is evident that this condition is answered, by supposing the product which we have just now found equal to 0: but this happens, not only when the first