Leonhard Euler · 1770 (1828 English ed.) · Internet Archive
Elements of Algebra
All 627 leaves · leaf 302
Transcription
CHAP. XII. OF ALGEBRA. 265
744. It often happens, however, that though such an equation has a rational root, that root cannot be found by the rule which we are now considering.
Let there be given the equation $x^3 = 6x + 40$, in which $x = 4$ is one of the roots. We have here $f = 6$ and $g = 40$; farther, $g^2 = 1600$, and $\frac{4}{27}f^3 = 32$; so that $g^2 - \frac{4}{27}f^3 = 1568$, and $\sqrt{(g^2 - \frac{4}{27}f^3)} = \sqrt{1568} = \ldots$ $\sqrt{(4 \cdot 4 \cdot 49 \cdot 2)} = 28 \sqrt{2}$; consequently one of the roots will be
$x = \sqrt[3]{(\frac{40 + 28 \sqrt{2}}{2})} + \sqrt[3]{(\frac{40 - 28 \sqrt{2}}{2})}$ or
$x = \sqrt[3]{(20 + 14 \sqrt{2})} + \sqrt[3]{(20 - 14 \sqrt{2})}$;
which quantity is really $= 4$, although, upon inspection, we should not suppose it. In fact, the cube of $2 + \sqrt{2}$ being $20 + 14 \sqrt{2}$, we have, reciprocally, the cube root of $20 + 14 \sqrt{2}$ equal to $2 + \sqrt{2}$; in the same manner, $\sqrt[3]{(20 - 14 \sqrt{2})} = 2 - \sqrt{2}$; wherefore our root $x = 2 + \sqrt{2} + 2 - \sqrt{2} = 4$*.
745. To this rule it might be objected, that it does not extend to all equations of the third degree, because the square of $x$ does not occur in it; that is to say, the second term of the equation is wanting. But we may remark, that every complete equation may be transformed into another, in which the second term is wanting, which will therefore enable us to apply the rule.
To prove this, let us take the complete equation $x^3 - 6x^2 + 11x - 6 = 0$: where, if we take the third of the coefficient 6 of the second term, and make $x - 2 = y$, we shall have
$x = y + 2$, $x^2 = y^2 + 4y + 4$, and $x^3 = y^3 + 6y^2 + 12y + 8$
Consequently, $-6x^2 = -6y^2 - 24y - 24$
$11x = 11y + 22$
$-6 = -6$
or, $x^3 - 6x^2 + 11x - 6 = y^3$ * $- y$ *
We have, therefore, the equation $y^3 - y = 0$, the resolu-
* We have no general rules for extracting the cube root of these binomials, as we have for the square root; those that have been given by various authors, all lead to a mixt equation of the third degree similar to the one proposed. However, when the extraction of the cube root is possible, the sum of the two radicals which represent the root of the equation, always becomes rational; so that we may find it immediately by the method explained, Art. 722. F. T.
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