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Find all $x$ such that $$x^6=(x+1)^6.$$

So far, I have found the real solution $x= -\frac{1}{2}$, and the complex solution $x = -\sqrt[3]{-1}$.

Are there more, and if so what/how would be the most efficient find all the solutions to this problem? I am struggling to find the rest of the solutions.

4 Answers 4

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Since $0$ is not a solution, this is equivalent to solve $$ \frac{x+1}{x}=\zeta $$ where $\zeta$ is any sixth root of $1$.

Then $x+1=\zeta x$, so $$ x=\frac{1}{\zeta-1} $$ Of course $\zeta=1$ should not be considered. There are other five sixth root of $1$, namely $$ e^{2k\pi i/3} $$ for $1\le k\le 5$.

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    The cleanest solution by far. – 2017-01-23
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    Fastest method, thank you very much. Can't believe it was so straightforward – 2017-01-23
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Hint We may factor this as:

$$(x-(x+1))(x+(x+1))(x^2-x(x+1)+(x+1)^2)(x^2+x(x+1)+(x+1)^2)=0$$

So then we have a linear and two quadratics which you should be able to solve.

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We can simplify the problem by substituting $x=y-\frac{1}{2}$.

$$x^6 = \left(x+1\right)^6$$

$$\left(y-\frac{1}{2}\right)^6 = \left(y+\frac{1}{2}\right)^6$$

If we expand both sides and then collect the terms, the even powers of $y$ drop out and only the odd powers remain.

$$6y^5+5y^3+\frac{3}{8}y=0$$

You can factor out the root at $y=0$ (i.e. corresponding to the solution $x=-\frac{1}{2}$) and then factor the remaining polynomial into two quadratics.

$$6y\left(y^2+\frac{3}{4}\right)\left(y^2+\frac{1}{12}\right)=0$$

Think you can take it from here?

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    Not sure why, but this means that 0 is a real solution? I am pretty sure that is not true in the original problem. – 2017-01-23
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    @user405892 No, $y=0$, which means $\displaystyle x = -{1\over2}$. – 2017-01-23
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Expand out the right hand side (binomial theorem will save you time): $$x^6=x^6+6x^5+15x^4+20x^3+15x^2+6x+1$$ Subtract $x^6$ from both sides: $$0=6x^5+15x^4+20x^3+15x^2+6x+1$$ Now just find the roots of this $5$th degree polynomial. If you are working over the complex numbers there will be exactly $5$ (see fundamental theorem of algebra), if you are only working in the real numbers $x=-\frac 12$ is the only solution.

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    How would you find the roots of the 5th degree polynomial? – 2017-01-23
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    Note that it is easier to subtract $(x+1)^6$ and factor via difference of squares. – 2017-01-23
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    @ZacharySelk I would start with rational roots theorem, which would give you a quartic. At this point it gets a lot tougher, but it is entirely possible to solve now using the Quartic Equation. Of course, factoring is a whole lot better. – 2017-01-23
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    @ZacharySelk OP already knows that $x= -1/2$ is a root so this can be easily factored into a quartic. This may not be the most efficient way to solve the problem but it's certainly correct. – 2017-01-23