Elements of Algebra

250 ELEMENTS SECT. IV. factor $x-2=0$, which gives us $x=2$, but also when the second factor $x^2+2x+4=0$. Let us, therefore, make $x^2+2x+4=0$; then we shall have $x^2 = -2x-4$, and thence $x = -1 \pm \sqrt{-3}$. 711. So that beside the case, in which $x=2$, which corresponds to the equation $x^3=8$, we have two other values of $x$, the cubes of which are also $8$; and these are, $x = -1+\sqrt{-3}$, and $x = -1-\sqrt{-3}$, as will be evident, by actually cubing these expressions; $-1+\sqrt{-3}$ \qquad\qquad $-1-\sqrt{-3}$ $-1+\sqrt{-3}$ \qquad\qquad $-1-\sqrt{-3}$ --- \qquad\qquad\qquad --- $1-\sqrt{-3}$ \qquad\qquad $1+\sqrt{-3}$ $-\sqrt{-3}-3$ \qquad\qquad $+\sqrt{-3}-3$ --- \qquad\qquad\qquad --- $-2-2\sqrt{-3}$ \quad square \quad $-2+2\sqrt{-3}$ $-1+\sqrt{-3}$ \qquad\qquad $-1-\sqrt{-3}$ --- \qquad\qquad\qquad --- $2+2\sqrt{-3}$ \qquad\qquad $2-2\sqrt{-3}$ $-2\sqrt{-3}+6$ \qquad\qquad $+2\sqrt{-3}+6$ --- \qquad\qquad\qquad --- $8.$ \qquad\qquad cube. \qquad\qquad $8.$ It is true, that these values are imaginary, or impossible; but yet they deserve attention. 712. What we have said applies in general to every cubic equation, such as $x^3=a$; namely, that beside the value $x=\sqrt[3]{a}$, we shall always find two other values. To abridge the calculation, let us suppose $\sqrt[3]{a}=c$, so that $a=c^3$, our equation will then assume this form, $x^3-c^3=0$, which will be divisible by $x-c$, as the actual division shews: $x-c) \ x^3 - c^3 \ (x^2 + cx + c^2$ \qquad $\underline{x^3 - cx^2}$ \qquad\qquad $cx^2 - c^3$ \qquad\qquad $\underline{cx^2 - c^2x}$ \qquad\qquad\qquad $c^2x - c^3$ \qquad\qquad\qquad $\underline{c^2x - c^3}$ \qquad\qquad\qquad\qquad $0.$ Consequently, the equation in question may be represented by the product $(x-c) \times (x^2+cx+c^2)=0$, which is in fact $=0$, not only when $x-c=0$, or $x=c$, but also