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Problem:Find the number of distinct real roots of $(x-a)^3+(x-b)^3+(x-c)^3=0$ where $a,b,c$ are distinct real numbers

Solution:$(x-a)^3+(x-b)^3+(x-c)^3=0$

$3x^3-3x^2(a+b+c)+3x(a^2+b^2+c^2)-a^3-b^3-c^3=0$

By Descartes rule of sign,number of positive real roots $=3$

But are they distinct $?$

Answer :- number of distinct real roots $ =1$

Community
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rst
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    Descartes' rule of signs doesn't tell you that - it tells you whether the number of positive roots is even or odd. Also, $a+b+c$ and $a^3 + b^3 + c^3$ could be negative or positive depending on $a,b,c$ – Cocopuffs Sep 21 '13 at 05:55
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    Descartes Rule of signs does not answer the question, for (i) it gives an upper bound on the number of positive roots, not the number and (ii) we cannot count sign changes by looking at $+$ and $-$ signs, some coefficients may not have the sign they appear to have. Also, the answer is not $3$. – André Nicolas Sep 21 '13 at 05:56
  • This isn't really a Number Theory question. Do you know about Rolle's Theorem from Calculus? – Gerry Myerson Sep 21 '13 at 05:56
  • I know Rolle's Theorem.But don't know "how to solve this question with it" – rst Sep 21 '13 at 06:00
  • @Cocopuffs @ André Nicolas, I got your points.But how can we solve this question – rst Sep 21 '13 at 06:16
  • What does Rolle's Theorem imply about what happens to the derivative of a (differentiable) function between two real zeros of the function? – Gerry Myerson Sep 21 '13 at 06:20
  • the derivative of a (differentiable) function between two real zeros of the function vanishes – rst Sep 21 '13 at 06:31
  • Your use of Descartes' Rule may be wrong: if $;a,b,c<0;$ , then the cubic has no change of signs... – DonAntonio Sep 21 '13 at 09:49

3 Answers3

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If $$f(x)=(x-a)^3+(x-b)^3+(x-c)^3$$ then $$f'(x)=3(x-a)^2+3(x-b)^2+3(x-c)^2$$ Since $a,b,c$ are distinct real numbers $f'(x) > 0$ for all $x\in\mathbb{R}$ and therefore $f$ is strictly increasing and therefore it has only one real root.

EDIT: The last statement is true since $f$ is a polynomial function of degree $3$ ($a_0>0$) so $\lim_{x\to-\infty} f(x) = -\infty$, $\lim_{x\to\infty} f(x) = \infty$ and $f$ is continuous.

kedrigern
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  • This means,every strictly increasing function has only one real root – rst Oct 07 '13 at 14:25
  • @ kedrigern, you might want to include that $\lim_{x\to\pm\infty}f(x)=\pm\infty$. – JSCB Oct 07 '13 at 14:29
  • Since $f$ is a polynomial function of degree $3$, I assumed the rest was clear. – kedrigern Oct 07 '13 at 14:35
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    Every strictly increasing function has at most one real root. In this case it is exactly one since (as J A S O N pointed out) $\lim_{x\to-\infty}f(x)=-\infty$, $\lim_{x\to\infty}f(x)=\infty$ and $f$ is continuous. – kedrigern Oct 07 '13 at 14:41
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Set $m=(a+b+c)/3$, $A=a-m$, $B=b-m$, $C=c-m$ and $x=y+m$. Then your equation becomes $$ (y-A)^3 + (y-B)^3 + (y-C)^3 = 0 $$ and, since $A+B+C=0$, your expansion applies to give $$ y^3+(A^2+B^2+C^2)-\frac{A^3+B^3+C^3}{3}=0 $$ which is a suppressed cubic, whose discriminant is $$ \frac{1}{4}\biggl(-\frac{A^3+B^3+C^3}{3}\biggr)^2+\frac{1}{27}(A^2+B^2+C^2)^3>0 $$ so the equation has only one real root.

egreg
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Let's substitute the variable: $x=y+\frac{(a+b+c)}{3}$

The equation will now look like

$$3y^3+(2a^2-2ab-2ac+2b^2-2bc+2c^2)y+\frac{(a+b-2c)(a-3b+c)(b-2a+c)}{9}=0$$

Now we apply Cardano's method.

$$Q=\left(\frac{2a^2-2ab-2ac+2b^2-2bc+2c^2}{3}\right)^3+\left(\frac{\frac{(a+b-2c)(a-3b+c)(b-2a+c)}{9}}{2}\right)^2=\frac{8}{27}(a^2-ab-ac+b^2-bc+c^2)^3+\frac{1}{324}(a+b-2c)^2(a-2b+c)^2(b-2a+c)^2$$

Then the answer is given depending on the sign of Q.

If $Q>0$ there is only one real root

If $Q<0$ there are three distinct real roots

If $Q=0$ there are two distinct real roots.

Тимофей Ломоносов
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