Elements of Algebra

450 ELEMENTS PART II. $z³ - 3z² + 3z - 1 = c$. If we should here, according to the rules already given, suppose the cube root to be $z-u$, and, by comparing the formula with the cube $z³ - 3uz² + 3u²z - u³$, determine $u$ so, that the second term may also vanish, we should have $u=1$; and the other terms forming the equation $3z = 3u²z - u³ = 3z - 1$, we should find $z = ∞$, from which we can draw no conclusion. Let us therefore rather leave $u$ undetermined, and deduce $z$ from the quadratic equation $-3z² + 3z = -3uz² + 3u²z - u³$, or $3uz² - 3z² = 3u²z - 3z - u³$, or $3(u-1)z² = 3(u²-1)z - u³$, or $z² = (u+1)z - \frac{u³}{3(u-1)}$; from this we shall find $z = \frac{u+1}{2} \pm \sqrt{(\frac{u²+2u+1}{4} - \frac{u³}{3(u-1)})}$ or $z = \frac{u+1}{2} + \sqrt{(\frac{-u³+3u²-3u-3}{12(u-1)})}$; so that the question is reduced to transforming the fraction under the radical sign into a square. For this purpose, let us first multiply the two terms by $3(u-1)$, in order that the denominator becoming a square, namely, $36(u-1)²$, we may only have to consider the numerator $-3u⁴ + 12u³ - 18u² + 9$: and, as the last term is a square, we shall suppose the formula, according to the rule, equal to the square of $gu² + fu + 3$, that is, to $g²u⁴ + 2fgu³ + f²u² + 6gu² + 6fu + 9$. We may make the last three terms disappear, by putting $6f=0$, or $f=0$, and $6g+f²=-18$, or $g=-3$; and the remaining equation, namely, $-3u + 12 = g²u + 2fu = 9u$, will give $u=1$. But from this value we learn nothing; so that we shall proceed by writing $u=1+t$. Now, as our formula becomes in this case $-12t - 3t⁴$, which cannot be a square, unless $t$ be negative, let us at once make $t=-s$; by these means we have the formula $12s - 3s⁴$, which becomes a square in the case of $s=1$. But here we are stopped again; for when $s=1$, we have $t=-1$, and $u=0$, from which we can draw no conclusion, except that in whatever manner we set about it, we shall never find a value that will bring us to the end proposed; and hence we may already infer, with some degree of certainty, that it is impossible to find two cubes whose sum is a cube. But we shall be fully convinced of this from the following demonstration. 243. *Theorem*. It is impossible to find any two cubes, whose sum, or difference, is a cube.