The first thing you do is "reduce" the equation to the form x^3+ bx= c without any x^2 term. To do that, let x= y- a. Then x^3= (y- a)^3= y^3- 3ay^2+ 3a^2y- a^3[/math] and x^2= (y- a)^2= y^2- 2ay+ a^2[/math]<br />
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x^3- 6x^2+ 2x-1= y^3 -ay^2+ a^2y- a^3- 6y^2+ 12ay- 6a^2+ 2y- 2a- 1= y^3+ (-a- 6)y^2+ (a^2+ 2)y+ (-a^3- 2a- 1)[/math]&lt;br /&gt;
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That will have no &amp;quot;y^2&amp;quot; term is a= -6 and, in that case, the polynomial is &lt;br /&gt;
y^3+ 38y+ 229 and so our equation is y^3+ 38y+ 229= 0.&lt;br /&gt;
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Here&amp;#039;s a quick review of Cardano&amp;#039;s formula:&lt;br /&gt;
(a+ b)^3= a^3+ 3a^2b+ 3ab^2+ b^3&lt;br /&gt;
-3ab(a+ b)= -3a^2b- 3ab^3&lt;br /&gt;
so that (a+ b)^3+ 3ab(a+ b)= a^3+ b^3.&lt;br /&gt;
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In particular, if we let x= a+b, m= 3ab, and n= a^3+ b^3, x^3+ mx= n.&lt;br /&gt;
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Can we go the other way? That is, given m and n, can we find a and b and so find x?&lt;br /&gt;
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Yes, we can. From m= 3ab, we have b= m/(3a) so a^3+ b^3= a^3+ m^3/(3^2a^3)= n. Multiplying through by a^3, (a^3)^2+ m^3/3^3= na^3 or (a^3)^2- na^3+ m^3/3^3= 0. &lt;br /&gt;
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We can think of that as a &lt;b&gt;quadratic&lt;/b&gt; equation in a^3 and solve it with the quadratic formula:&lt;br /&gt;
a^3= \frac{n\pm\sqrt{n^2- 4\frac{m^3}{3^3}}}{2}= \frac{n}{2}\pm\sqrt{\left(\frac{n}{2}\right)^2- \left(\frac{m}{3}\right)^3}&lt;br /&gt;
and a is the cube root of that.&lt;br /&gt;
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From n= a^3+ b^3 we have &lt;br /&gt;
b^3= n- a^3= \frac{n}{2}\mp\sqrt{\left(\frac{n}{2}\right)^2- \left(\frac{m}{3}\right)^3}.&lt;br /&gt;
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Now, in this problem, y^3+ 38y+ 229= 0 or y^3+ 38y= -229 so m= 38 and n= -229. &lt;br /&gt;
\frac{n}{2}= -\frac{229}{2} &lt;br /&gt;
and &lt;br /&gt;
\left(\frac{n}{2}\right)^2= \frac{52441}{4}&lt;br /&gt;
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\frac{m}{3}= \frac{38}{3}&lt;br /&gt;
and&lt;br /&gt;
\left(\frac{m}{3}\right)^3= \frac{54872}{27}&lt;br /&gt;
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So &lt;br /&gt;
a^3= -\frac{229}{2}\pm\sqrt{\frac{52441}{4}- \frac{54872}{27}}&lt;br /&gt;
and&lt;br /&gt;
b^3= -\frac{229}{2}\mp\sqrt{\frac{52441}{4}- \frac{54872}{27}}&lt;br /&gt;
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Calculate those numbers, take the cube roots to find a and b and then find x= a+ b.&lt;br /&gt;
Each of those will have 3 cube roots but in the various ways of combining them, some things will cancel so that there will be, at most, 3 roots to the equation.
