Elements of Algebra

380 ELEMENTS PART II. diately perceive the known case to be that of $x = 0$. The second class comprehends the formula $a + bx + cx^2 + g^3x^3$; that is to say, the case in which the last term is a cube. The third class is composed of the two former, and comprehends the cases in which both the first term and the last are cubes. 149. Case 1. Let $f^3 + bx + cx^2 + dx^3$ be the proposed formula, which is to be transformed into a cube. Suppose its root to be $f + px$; and, consequently, that the formula itself is equal to the cube, $f^3 + 3f^2px + 3fp^2x^2 + p^3x^3$; as the first terms disappear of themselves, we shall determine $p$, so as to make the second terms disappear also; namely, by making $b = 3f^2p$, or $p = \frac{b}{3f^2}$; then the remaining terms being divided by $x^2$, give $c + dx = 3fp^2 + p^3x$; or $x = \frac{c-3fp^2}{p^3-d}$. If the last term, $dx^3$, had not been in the formula, we might have simply supposed the cube root to be $f$, and should have then had $f^3 = f^3 + bx + cx^2$, or $b + cx = 0$, and $x = -\frac{b}{c}$; but this value would not have served to find others. 150. Case 2. If, in the second place, the proposed expression have this form, $a + bx + cx^2 + g^3x^3$, we may represent its cube root by $p + gx$, the cube of which is $p^3 + 3p^2gx + 3gp^2x^2 + g^3x^3$; so that the last terms destroy each other. Let us now determine $p$, so that the last terms but one may likewise disappear; which will be done by supposing $c = 3g^2p$, or $p = \frac{c}{3g^2}$, and the other terms will then give $a + bx = p^3 + 3gp^2x$; whence we find $x = \frac{a-p^3}{3gp^2-b}$. If the first term, $a$, had been wanting, we should have contented ourselves with expressing the cube root by $gx$, and should have had $g^3x^3 = bx + cx^2 + g^3x^3$, or $b + cx = 0$,