Here, we suppose that $x,y\in\mathbb{R}$ and that $x^n=y^n$, where $n$ is odd. I want to prove that $x=y$.
Maybe we can use that $x^n-y^n=(x-y)(x^{n-1}+x^{n-2}y+...+xy^{n-2}+y^{n-1})$
So, it suffices to show that if $x^n=y^n$ then $x^{n-1}+x^{n-2}y+...+xy^{n-2}+y^{n-1}\neq 0$
Any hint?
Since $n$ is odd, $x^n=y^n$ implies that $x$ and $y$ have the same sign. We may therefore assume that they are both strictly positive. But then $x^{n-1}+x^{n-2}y+\cdots+xy^{n-2}+y^{n-1}$ is strictly positive and so cannot be zero.
Since this is tagged "calculus" you may want to show that $f(x) = x^n$ is increasing using the derivative of $f$.
EDIT: the tag has since been changed.
Hint: Verify that for $x=0$, $y=0$ is the only solution and trivially $x=y$ for this case. Now let $z = \frac yx, x \neq 0$.
Then $y^n - x^n = x^n(z^n - 1)$
Zeroes of the LHS are zeroes of the RHS.
Meaning that $z^n - 1 = 0$.
Consider the $n$-th roots of unity for odd $n$.
Can you continue from there?
A Proof by Contradiction:
Suppose $ x \neq y$. Then $ x \gt y $ or $x \lt y$. Consider the first case. It won't hurt to suppose that $x , y \gt 0$. I shall explain a little later. Now let us prove the statement $ x^n \gt y^n $ by induction on $n$. Base case is our assumption. Suppose $ x^n \gt y^n $. Since we know that $x \gt y \gt 0$ we can multiply the two inequalities to obtain $x^{n + 1} \gt y^{n + 1}$. Hence we have that $$ x^n \gt y^n \;\; \forall n \in \Bbb N $$
Hence $ x^n = y^n $ for some odd $n \in \Bbb N$ is impossible leading to a contradiction.
Now the second case: $ x \lt y$ is almost identical.
Now what if the signs of $x$ and $y$ were different. Again without loss of generality assume $ x \lt 0 \lt y $. Now prove by induction that $ x^n = (-1)^n|x|^n $. Then for odd $n$ we have that $ x^n = (-1)^n|x|^n \lt 0 \lt y^n$ leading to a contradiction. The case for $ x \gt 0 \gt y $ is again identical.
One more case, if $x = 0$ and $y \neq 0$. Prove by induction that $y^n \neq 0$ for each $n \Bbb N$. But then $x^n = 0$ leads to a contradiction. The case for $ y = 0 $ and $x \neq 0$ is almost identical.
All cases are exhausted (I think). Didn't think it would be this long when I started writing.
