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I have this exercise where I have no idea how to proceed: I need to find the number of solutions for x in $x^3\:−\:x\:=\:a^3\:−\:a$
Any clues? Tried $x^3\:−\:x\:-a^3\:+\:a=\:0$ but I still can't proceed. I would like to factor it so it can be a second degree polynomial, and then I can use $Δ = b^2 - 4ac$

nginx9101
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  • You have exactly 3 solutions because you have a polynomial of degree 3. You can also try to work with the expression, $x^3-a^3 = x-a$. You see inmediatly that $x=a$ is a solution and then factorize $x^3 - a^3$ to simplify your work. – thewatcher Aug 10 '20 at 17:01
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    If you're trying to count real solutions, then calculate the the cubic discriminant, instead of the quadratic discriminant. – Mars Aug 10 '20 at 17:01
  • You can see that $x=a$ is a root for the equation, then you can easily factor it out no need for cubic discriminants – Anindya Prithvi Aug 10 '20 at 17:04
  • @Morph I still haven't covered cubic polynomials, would prefer to use the factor solution – nginx9101 Aug 10 '20 at 17:17

3 Answers3

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You want to solve $f(x) = x^3-x-a^3+a = 0$

$f(x) = (x^3-a^3)-(x-a) = (x-a)(x^2+ax+a^2) - (x-a) = (x-a)(x^2+ax+a^2-1)$.

So clearly $x = a$ is a solution. Now can you use the discriminant on the quadratic term?

mihirb
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  • Yes, now I can proceed. However, how did you find $(x^2+ax+a^2)$ ? I wouldn't have thought of it on my own – nginx9101 Aug 10 '20 at 17:16
  • @nginx9101 $a^3-b^3 = (a-b)(a^2+ab+b^2)$ so I just factored $x^3-a^3$ using that. It is a common factorization I think you should memorize. $a^3-b^3 = (a-b)(a^2+ab+b^2)$ and $a^3+b^3 = (a+b)(a^2-ab+b^2)$. – mihirb Aug 10 '20 at 17:18
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enter image description here

You can use the graph. Consider that the green graph is for $p(x)=x^3-x$ and the horizontal lines are $a^3-a$. The number of intersections is the number of solutions that you are looking for.

Can you see the cases you have to analyze?

Arnaldo
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  • Three intersections indeed – nginx9101 Aug 10 '20 at 17:24
  • @nginx9101 note there are also cases with two intersections (not just 3 and 1) if you make the line exactly tangent to one of the maxima or minima of the graph. that's a little trickier to see. – mihirb Aug 10 '20 at 17:35
  • @nginx9101: You can have one, two or three intersections. – Arnaldo Aug 10 '20 at 20:37
  • @Arnaldo How can I exploit this? I don't get it – nginx9101 Aug 10 '20 at 22:35
  • @nginx9101: each straight line ($g,p,h$) represents a value of $a^3-a$. The function $f$ is $f(x)=x^3-x$. So, the number of solutions will be the number of intersections. If you vary the value of $a$ you vary the horizontal line and then the number of solutions. – Arnaldo Aug 11 '20 at 11:55
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Hint: Obviously $x=a$ is a solution. Thus factoring out $(x-a)$ gives $$x^3-a^3-x+a=(x-a)(x^2+ax+a^2-1)=0.$$ Can you take it from here?

b00n heT
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