If your students know a little about the Eisenstein integers (unique factorization and what the units are) there's the following simple argument (maybe it's essentially Euler's?). Let $u$ and $v$ be the complex roots of $z^2+z+1=0$.

Theorem: Let $A$, $B$, $C$ be non-zero elements of $Q[u]$ with sum $0$ and product twice a cube. Then some two of $A$, $B$, $C$ are equal.

Corollary: Suppose $x$ is in $Q$ and $x^2-1$ is a cube. Then $x$ is $1$, $-1$, $0$, $3$, or $-3$.

(To prove the corollary let $A=1+x$, $B=1-x$ and $C=-2$).

The proof of the theorem is a reductio. If there's a counterexample, there's one with $A$, $B$, $C$ in $Z[u]$; take such a counterexample with $d=\min(/A/,/B/,/C/)$ as small as possible. Then $A$, $B$ and $C$ are pairwise prime. Since $ABC=2$(cube) we may assume $A=2i$(cube), $B=j$(cube) $C=k$(cube) where $i$, $j$, and $k$ are in the set $\{1,u,v\}$. Now all cubes in $Z[u]$ are $0$ or $1$ mod $2$. Since $B+C$ is $0$ mod $2$, $j=k$. Since $ABC=2$(cube), $ijk$ is a cube and $i=j=k$. We may assume $i=j=k=1$. Then $A=2r^3$, $B=s^3$, $C=t^3$, and we may further assume that $s$ and $t$ are $1$ mod $2$. $s$ and $t$ are not both in $\{1,-1\}$ and it follows that $d$ is at least $\sqrt{27}$. Now look at $s+t$, $us+vt$ and $vs+ut$. They sum to $0$ and their product is $B+C=-2(r^3)$. They are congruent to $0$, $1$ and $1$ mod $2$, and the last 2 of them can't be equal since $s$ is not equal to $t$. Since each of them is at most $2d^{1/3}$, this contradicts the minimality assumption.

This is really a 3-descent argument on an elliptic curve, but the fancy language as you see isn't needed.An almost identical argument gives what I think is the nicest proof of FLT for exponent 3.

If your students know a little about the Eisenstein integers (unique factorization and what the units are) there's the following simple argument (maybe it's essentially Euler's?). Let $u$ and $v$ be the complex roots of $z^2+z+1=0$.

Theorem. Let $A$, $B$, $C$ be non-zero elements of $\mathbb{Q}[u]$ with sum $0$ and product twice a cube. Then some two of $A$, $B$, $C$ are equal.

Corollary. Suppose $x$ is in $\mathbb{Q}$ and $x^2-1$ is a cube. Then $x$ is $1$, $-1$, $0$, $3$, or $-3$.

(To prove the corollary let $A=1+x$, $B=1-x$ and $C=-2$).

The proof of the theorem is a reductio ad absurdum. If there's a counterexample, there's one with $A$, $B$, $C$ in $\mathbb{Z}[u]$; take such a counterexample with $d=\min(|A|,|B|,|C|)$ as small as possible. Then $A$, $B$, $C$ are pairwise coprime. Since $ABC=2$(cube) we may assume $A=2i$(cube), $B=j$(cube), $C=k$(cube) where $i$, $j$, and $k$ are in the set $\{1,u,v\}$. Now all cubes in $\mathbb{Z}[u]$ are $0$ or $1$ mod $2$. Since $B+C$ is $0$ mod $2$, $j=k$. Since $ABC=2$(cube), $ijk$ is a cube and $i=j=k$. We may assume $i=j=k=1$. Then $A=2r^3$, $B=s^3$, $C=t^3$, and we may further assume that $s$ and $t$ are $1$ mod $2$. $s$ and $t$ are not both in $\{1,-1\}$ and it follows that $d$ is at least $\sqrt{27}$. Now look at $s+t$, $us+vt$ and $vs+ut$. They sum to $0$ and their product is $B+C=-2(r^3)$. They are congruent to $0$, $1$ and $1$ mod $2$, and the last 2 of them can't be equal since $s$ is not equal to $t$. Since each of them is at most $2d^{1/3}$, this contradicts the minimality assumption.

This is really a 3-descent argument on an elliptic curve, but the fancy language as you see isn't needed. An almost identical argument gives what I think is the nicest proof of Fermat's Last Theorem for exponent 3.

If your students know a little about the Eisenstein integers (unique factorization and what the units are) there's the following simple argument (maybe it's essentially Euler's?). Let u and v be the complex roots of z^2+z+1=0.

Theorem: Let A,B,C be non-zero elements of Q[u] with sum 0 and product twice a cube. Then some two of A,B,C are equal.

Corollary: Suppose x is in Q and x^2-1 is a cube. Then x is 1,-1,0,3, or -3.

(To prove the corollary let A=1+x, B=1-x and C=-2).

The proof of the theorem is a reductio. If there's a counterexample, there's one with A,B,C in Z[u]; take such a counterexample with d=min(/A/,/B/,/C/) as small as possible. Then A,B and C are pairwise prime. Since ABC=2(cube) we may assume A=2i(cube), B=j(cube) C=k(cube) where i,j,and k are in the set {1,u,v}. Now all cubes in Z[u] are 0 or 1 mod 2. Since B+C is 0 mod 2, j=k. Since ABC=2(cube), ijk is a cube and i=j=k. We may assume i=j=k=1. Then A=2r^3,B=s^3,C=t^3, and we may further assume that s and t are 1 mod 2. s and t are not both in {1,-1} and it follows that d is at least root(27). Now look at s+t, us+vt and vs+ut. They sum to 0 and their product is B+C=-2(r^3). They are congruent to 0,1 and 1 mod 2, and the last 2 of them can't be equal since s is not equal to t. Since each of them is at most 2(d^(1/3), this contradicts the minimality assumption.

This is really a 3-descent argument on an elliptic curve, but the fancy language as you see isn't needed.An almost identical argument gives what I think is the nicest proof of FLT for exponent 3.

If your students know a little about the Eisenstein integers (unique factorization and what the units are) there's the following simple argument (maybe it's essentially Euler's?). Let $u$ and $v$ be the complex roots of $z^2+z+1=0$.

Theorem: Let $A$, $B$, $C$ be non-zero elements of $Q[u]$ with sum $0$ and product twice a cube. Then some two of $A$, $B$, $C$ are equal.

Corollary: Suppose $x$ is in $Q$ and $x^2-1$ is a cube. Then $x$ is $1$, $-1$, $0$, $3$, or $-3$.

(To prove the corollary let $A=1+x$, $B=1-x$ and $C=-2$).

The proof of the theorem is a reductio. If there's a counterexample, there's one with $A$, $B$, $C$ in $Z[u]$; take such a counterexample with $d=\min(/A/,/B/,/C/)$ as small as possible. Then $A$, $B$ and $C$ are pairwise prime. Since $ABC=2$(cube) we may assume $A=2i$(cube), $B=j$(cube) $C=k$(cube) where $i$, $j$, and $k$ are in the set $\{1,u,v\}$. Now all cubes in $Z[u]$ are $0$ or $1$ mod $2$. Since $B+C$ is $0$ mod $2$, $j=k$. Since $ABC=2$(cube), $ijk$ is a cube and $i=j=k$. We may assume $i=j=k=1$. Then $A=2r^3$, $B=s^3$, $C=t^3$, and we may further assume that $s$ and $t$ are $1$ mod $2$. $s$ and $t$ are not both in $\{1,-1\}$ and it follows that $d$ is at least $\sqrt{27}$. Now look at $s+t$, $us+vt$ and $vs+ut$. They sum to $0$ and their product is $B+C=-2(r^3)$. They are congruent to $0$, $1$ and $1$ mod $2$, and the last 2 of them can't be equal since $s$ is not equal to $t$. Since each of them is at most $2d^{1/3}$, this contradicts the minimality assumption.

This is really a 3-descent argument on an elliptic curve, but the fancy language as you see isn't needed.An almost identical argument gives what I think is the nicest proof of FLT for exponent 3.