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Find roots of: $$x^{6}\ -\ \left(x-1\right)^{6}=0 \tag {1}$$

I know this equation has $4$ complex roots and exactly one real roots of value $0.5$.

However, my first instinct was to do this: $$x^{6}\ =\ \left(x-1\right)^{6} \tag{2}$$ "raise both sides to 6-th power" to get: $$x=x-1\tag{3}$$

Which has no real solution. I see that this wrong. How to avoid this error? Thanks.

Inspired by watching this youtube video

Edit:

I am not asking about how to solve the problem. I want to know what I did wrong from an Algebraic stand-point. Maybe raising to the power? What is wrong with that?

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    $\begingroup$ You meant "raise both sides to the $1/6$ power". Since $6$ is even, for $x$ real it is $x^6=(x-1)^6 \iff |x|=|x-1|$. $\endgroup$
    – Bernkastel
    Feb 16 at 23:49
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    $\begingroup$ @NoChance Note that $x^6 - (x-1)^6 = 0$ is a polynomial of degree $5$ (since the $x^6$ terms cancel each other), so if there's $4$ complex roots, there can only be $1$ real root, with $x = 0.5$ being it (coming from $|x| = |x-1|$ mentioned in Bernkastel's comment). $\endgroup$
    – John Omielan
    Feb 16 at 23:54
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    $\begingroup$ @NoChance Haven't you read the comment of Bernkastel? You've forgotten the absolute value. $\endgroup$
    – callculus42
    Feb 17 at 0:13
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    $\begingroup$ What you did wrong is simply believing $a^6=b^6\implies a=b.$ $\endgroup$
    – Anne Bauval
    Feb 17 at 0:17
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    $\begingroup$ To amplify a bit on the comment of @AnneBauval, taking a sixth root (or any even root) in the real numbers is not necessarily a well defined function. That's because if $x^6=t$, it's necessarily also the case that $(-x)^6=t$ as well. By convention we usually say that the sixth root of a positive number is a positive number, but in a situation like yours, that choice does not exhaust all possibilities and the word the shouldn't be used because there are multiple possible sixth roots to consider. $\endgroup$
    – Robert Shore
    Feb 17 at 0:22

2 Answers 2

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Thanks for all the posted comments above. At the moment, no one had posted an answer, but I understood the following, which combined may provide an answer.

$$|x|=|x-1|$$

can't always be written as $x=x-1$. I need to learn how to solve such an equation.

Also, $$a^n=b^n$$

Does not always imply that $a=b$. The result is affected by the domain of $a$ and $b$ and whether the power is even or odd or integer or not, maybe among other factors.

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    $\begingroup$ Note for real $y$ and $z$ that $|y| = |z|$ means $y = z$ or $y = -z$. In your case with $|x| = |x-1|$, since $x \neq x - 1$, then we must instead have $x = -(x - 1) = -x + 1 \; \to \; 2x = 1 \; \to \; x = 0.5$. $\endgroup$
    – John Omielan
    Feb 17 at 1:03
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    $\begingroup$ @JohnOmielan, excellent explanation. Thank you. $\endgroup$
    – NoChance
    Feb 17 at 1:03
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If you had started with $x^2 - (x-1)^2=0$ then you can expand it to $x^2-x^2+2x-1=0$ and so $x=\frac12$.

But let's take a version of your approach. You might say $a^2=b^2$ clearly implies $a=b$ or $a=-b$; let's write this as $a=\omega b$ where $\omega^2=1$. We expect there to be be two possible values for $\omega$, namely $+1$ and $-1$. But if $a=x$ and $b=x-1$ you get $x=\omega(x-1)$ and so $x= \frac{-\omega}{1-\omega}= 1 - \frac{1}{1-\omega}$; when $\omega=-1$ it gives $x=\frac12$ as expected, while if $\omega=1$ it tries to give $1- \frac{1}{0}$ which is not a finite number, reflecting the fact there is no finite solution to $x=x-1$.

Translate this to your $6$th powers in $(2)$:

  • $a^6=b^6$ has solutions of the form $a=\omega b$ where $\omega^6=1$;
  • there are potentially six such $\omega$ (some complex) of which one is $\omega=1$
  • If $a=x$ and $b=x-1$ you get $x=\omega(x-1)$ and again $x= 1 - \frac{1}{1-\omega}$
  • So each of the six $\omega$ will give you a different solution for $x$, except $\omega=1$ which tries to give $1-\frac{1}{0}$ (still not a finite number).
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  • $\begingroup$ Thank you for taking the time to write such a clear explanation. $\endgroup$
    – NoChance
    Feb 17 at 1:10

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