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How many distinct real roots does $ (x^2 + x – 2)^3 + (8–2x^2 )^3 = (x^2 + 3x + 2)^3 $have?

If I didn't make any mistake that equation could be reduced to the form:$$(2+6x^2)(x+2)^3=(8–2x^2 )^3$$

However,I don't know how to approach next as more simplification and then forming up the factors again would be more time consuming,I suppose this problem could be solved in a more easy way,any ideas?

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I believe $(x+2)^3$ is factor of both sides (in the original and what you got), and cancelling that gives a quadratic (if I did the algebra right). –  Aryabhata Aug 27 '11 at 7:58
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Quadratic!? I got a cubic after the entire simplification following your hint:$4x^3-21x^2+48x-31=0$ (if I did the algebra right) –  Quixotic Aug 27 '11 at 8:37
    
@FoolForMath: Yes, but note that the cubic can be factored as follows: $$4x^{3}-21x^{2}+48x -31=\left( x-1\right) \left( 4x^{2}-17x+31\right).$$ –  Américo Tavares Aug 27 '11 at 11:20

4 Answers 4

up vote 2 down vote accepted

Using the idea in Aryabhata's comment, since

$$\left( 8-2x^{2}\right) ^{3}=2^{3}\left( 4-x^{2}\right) ^{3}=8 (2-x)^{3}(2+x)^{3},$$

your second equation

$$\left( 2+6x^{2}\right) \left( x+2\right) ^{3}=\left( 8-2x^{2}\right) ^{3}\tag{1}$$

is equivalent to

$$\begin{eqnarray*} \left( 2+6x^{2}\right) \left( x+2\right) ^{3} &=&8( 2-x)^{3}(2+x) ^{3}.\tag{2} \end{eqnarray*}$$

So the factor $(x+2)^3$ in both sides yields the real triple root $x=-2$. For $x\ne -2$ this equation $(2)$ is equivalent to

$$\begin{eqnarray*} \left( 2+6x^{2}\right) &=&-8\left( x-2\right) ^{3}\\ \left( 2+6x^{2}\right) +8\left( x-2\right) ^{3} &=&0 \\ \left( 1+3x^{2}\right) +4\left( x-2\right) ^{3} &=&0\text{.} \end{eqnarray*}$$

By inspection we see that $x=1$ is a root and by long division or Ruffini's rule, we find:

$$ \begin{eqnarray*} \left( 1+3x^{2}\right) +4\left( x-2\right) ^{3} =4x^{3}-21x^{2}+48x-31=\left( x-1\right) \left( 4x^{2}-17x+31\right). \end{eqnarray*}$$

Alternatively we could have applied the rational zero theorem: all the rational roots of the equation $$d_{n}x^{n}+d_{n-1}x^{n-1}+\ldots +d_{0}=0,$$ where all the coefficients are integers, are of the form $\frac{p}{q}=\frac{\text{a factor of }d_{0}}{\text{a factor of }d_{n}}$.

Hence, for $x\ne -2$, the equation $(1)$ is equivalent to

$$\left( x-1\right) \left( 4x^{2}-17x+31\right) =0.\tag{3}$$

Since the discriminant of the quadratic term is negative $$\Delta =17^{2}-4\times 4\times 31=-207<0,$$ the original equation has the real solution $x=1$ due to the factor $(x-1)$ and the triple real solution $x=-2$ due to the common factor $(x+2)^3$. In the total, two distinct real roots.

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+1,As always,there is lot more to learn from your answers :-) –  Quixotic Aug 27 '11 at 13:22
    
@FoolForMath, Thank you! –  Américo Tavares Aug 27 '11 at 13:28
    
No,thank you! :-) –  Quixotic Aug 27 '11 at 13:33

We are trying to solve the equation $$(x^2 + x – 2)^3 + (8–2x^2 )^3 -(x^2 + 3x + 2)^3=0.$$ Let $P(x)$ be the polynomial on the left-hand side. Note that $x^2+x-2=(x+2)(x-1)$, and $8-2x^2=-2(x-2)(x+2)$, and $x^2+3x+2=(x+2)(x+1).$ Thus $$P(x)=(x+2)^3[(x-1)^3-8(x-2)^3-(x+1)^3]=(x-2)^3Q(x).$$

Expand the remaining cubes. We get $$Q(x)=-8x^3+42x^2-96x+62.$$ We get very lucky: $x=1$ is a root of $Q(x)$. Divide. $$Q(x)=(x-1)(-8x^2+34x-62).$$ It is easy to verify, say by the Quadratic Formula, that $8x^2-34x+62=0$ has no real roots.

So there are altogether two distinct real roots, $x=-2$ and $x=1$.

Your calculation: Things would have worked out fine, if after the (correct) simplification of the difference of cubes you had used the fact that $8-2x^2=-2(x+2)(x-2)$. But except in special circumstances, of which this happens to be one, it is easier to deal with a polynomial equation of shape $P(x)=0$ than with one of shape $P_1(x)=P_2(x)$.

Comment: The question has a quite artificial character, because of the occurrence of an obvious root $x=-2$ of multiplicity at least $3$. Given an "arbitrary" degree $6$ polynomial, the techniques for finding the number of real roots would be quite different. Nowadays, an almost automatic first step is to see what graphing software, or a tool like Wolfram Alpha, has to say.

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+1 for Your calculation: and Comment: part! –  Quixotic Aug 27 '11 at 15:16

There's always Sturm's theorem. Sometimes it's better to spend a few minutes on brute force than a few hours looking for a clever solution.

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As this is a pre-algebra/precalc question, I don't think that this is a viable answer for the OP. But perhaps that's only because I'm used to using Sturm on the function and its derivative. Perhaps you use a different set? –  mixedmath Aug 27 '11 at 8:39

I am just a lazy, mathematician who happens to have access to Mathematica. So here you go:

NSolve[(x^2 + x - 2)^3 + (8 - 2 x^2)^3 == (x^2 + 3 x + 2)^3, x, Reals]

Out[7]= {{x->-2.},{x->-2.},{x->-2.},{x->1.}}

When you drop Reals, you get:

Out[8]= {{x->-2.},{x->-2.},{x->-2.},{x->1.},{x->2.125 -1.79844 I},{x->2.125 +1.79844 I}}

as an answer. But you were probably looking for a more intelligent way than an overgrown calculator.

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I just saw in your profile that you appearantly know Mathematica. Sorry in being this blunt.. –  Willem Noorduin Aug 27 '11 at 12:31
    
It's alright,yes I know Mathematica (not an expert but I am comfortable) and yes you are right I am looking for a intelligent manual solution that is how I am supposed to solve this during exams :-) –  Quixotic Aug 27 '11 at 13:35

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