How can I get help with solving cubic equations for my homework?

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Homework Statement

I was just wondering if anyone could give me a quick lesson on solving cubic equations(Ax^3+Bx^2+Cx+D=0), or a good place to go to, if anyone can help thanks alot



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The Attempt at a Solution

 
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here's the way I would argue it:

First, convert from the general cubic to the "reduced cubic" (without the x2 term). If we replace x by y- u, then
Ax^3+ Bx^2+ Cx+ D= A(y-u)^3+ B(y- u)^2+ C(y- u)+ D= Ay^3- 3Auy^2+ 3Au^2y- Au^3+ By^2- 2Buy+ Bu^2+ Cy- Cu+ D
= Ay^3+ (-3Au+ B)y^2+ (3Au^2- 2Bu+ C)y+ (-Au^3+ Bu^2- Cu)+ D[/itex]<br /> Choose u so that the coefficient of y<sup>2</sup> is 0: That is, choose u= B/3A so that the equation for y has no y<sup>2</sup> term. Solve that equation for y, and then x= y+ u.<br /> <br /> <br /> Now to solve that reduced equation:<br /> If a and b are any two numbers, then <br /> (a+b)^3= a^3+ 3a^2b+ 3ab^2+ b^3<br /> and<br /> 3ab(a+b)= 3a^2b+ 3ab^2<br /> so<br /> (a+ b)^3- 3ab(a+b)= a^3+ b^3<br /> or, letting x= a+b, m= 3ab, and n=a<sup>3</sup>+ b<sup>3</sup>, <br /> x^3- mx= n<br /> <br /> Given any &quot;reduced&quot; equation Ax<sup>3</sup>+ Bx+ C= 0, We can divide through by A to get x<sup>3</sup>+ (B/A)x+ C= 0 or x<sup>3</sup>- (-B/A)x= -C. m= -B/A and n= -C/A.<br /> <br /> Given m and n, can we solve for a and b, and so find x? <br /> <br /> Of course, we can! From m= 3ab, b= m/3a. Replacing b by that in n= a<sup>3</sup>+ b<sup>3</sup>, n= a<sup>3</sup>+ m<sup>3</sup>/(3<sup>3</sup>a<sup>3</sup>).l<br /> <br /> Multiply through by a<sup>3</sup> to get na<sup>3</sup>= a<sup>6</sup>+ m<sup>3</sup>/3<sup> which we can write as a quadratic equation for a<sup>3</sup>:<br /> (a<sup>3</sup>)<sup>3</sup>- na<sup>3</sup>+ m<sup>3</sup>/3<sup>3</sup> and solve that with the quadratic formula:<br /> <br /> a^3= \frac{n\pm\sqrt{n^2- 4m^2/3^3}}{2}= \frac{n}{2}\pm\sqrt{\left(\frac{n}{2}\right)^2-\left(\frac{m}{3}\right)^3}<br /> <br /> Of course, once you have found a, b= m/3a and x= a+ b.<br /> <br /> Using that is a heck of a lot of work which is why it is not normally taught in basic algebra!<br /> <br /> Here&#039;s a website where they just give the formula:<br /> <a href="http://www.math.vanderbilt.edu/~schectex/courses/cubic/" target="_blank" class="link link--external" rel="nofollow ugc noopener">http://www.math.vanderbilt.edu/~schectex/courses/cubic/</a></sup>
 
hmm that's a little bit more complicated than i thought but thank you
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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