Elements of Algebra

CHAP. X. OF ALGEBRA. 381 whence $x = -\frac{b}{c}$; but this is of no use for finding other values. 151. Case 3. Lastly, let the formula be, $f^3 + bx + cx^2 + g^3x^3$, in which the first and the last terms are both cubes. It is evident that we may consider this as belonging to either of the two preceding cases; and, consequently, that we may obtain two values of $x$. But beside this, we may also represent the root by $f+gx$, and then make the formula equal to the cube, $f^3 + 3f^2gx + 3fg^2x^2 + g^3x^3$; and likewise, as the first and last terms destroy each other, the others being divisible by $x$, we arrive at the equation $b+cx = 3f^2g + 3fg^2x$, which gives $x = \frac{b-3f^2g}{3fg^2-c}$. 152. On the contrary, when the given formula belongs not to any of the above three cases, we have no other resource than to try to find such a value for $x$ as will change it into a cube; then, having found such a value, for example, $x=h$, so that $a + bh + ch^2 + dh^3 = k^3$, we suppose $x = h + y$, and find, by substitution, $a$ $bh + by$ $ch^2 + 2chy + cy^2$ $dh^3 + 3dh^2y + 3dhy^2 + dy^3$ --- $k^3 + (b+2ch+3dh^2)y + (c+3dh)y^2 + dy^3$. This new formula belonging to the first case, we know how to determine $y$, and therefore shall find a new value of $x$, which may then be employed for finding other values. 153. Let us endeavour to illustrate this method by some examples. Suppose it were required to transform into a cube the formula $1+x+x^2$, which belongs to the first case. We might at once make the cube root 1, and should find $x+x^2=0$, that is, $x(1+x)=0$, and, consequently, either $x=0$, or $x=-1$; but from this we can draw no conclusion. Let us therefore represent the cube root by $1+px$; and as its cube is $1+3px+3p^2x^2+p^3x^3$, we shall have $3p=1$, or $p=\frac{1}{3}$; by which means the other