# Mistake in raising power

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?

• You meant "raise both sides to the $1/6$ power". Since $6$ is even, for $x$ real it is $x^6=(x-1)^6 \iff |x|=|x-1|$. Feb 16 at 23:49
• @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). Feb 16 at 23:54
• @NoChance Haven't you read the comment of Bernkastel? You've forgotten the absolute value. Feb 17 at 0:13
• What you did wrong is simply believing $a^6=b^6\implies a=b.$ Feb 17 at 0:17
• 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. Feb 17 at 0:22

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.

• 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 $x = -(x - 1) = -x + 1 \; \to \; 2x = 1 \; \to \; x = 0.5$. Feb 17 at 1:03
• @JohnOmielan, excellent explanation. Thank you. Feb 17 at 1:03

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).
• Thank you for taking the time to write such a clear explanation. Feb 17 at 1:10