By Lemma 1.31, we can conclude that ${\left[n\right]}^3=1\vee {\left[n\right]}^3=8$.

By hypothesis we know that $3\nmid abc$, which implies that $3\nmid a$, $3\nmid b$ and $3\nmid c$.
By repeatedly applying Lemma 3.1 for each case, we can conclude that

By hypothesis we have that $3\mid a^3+b^3=c^3$, which implies that $3\mid c$, from which we can conclude that $3\mid gcd(a,b,c)$.

By hypothesis we have that $3\mid c^3-a^3=b^3$, which implies that $3\mid b$, from which we can conclude that $3\mid gcd(a,b,c)$.

By hypothesis we have that $3\mid c^3-b^3=a^3$, which implies that $3\mid a$, from which we can conclude that $3\mid gcd(a,b,c)$.

By contradiction we assume that

By Lemma 1.30 we can assume that $gcd(a,b,c)=1$.
By Theorem 3.2 we can assume that $3\mid abc$, from which it follows that

We proceed by analysing each case:

• Case $3\mid a$.
  Let $a^{\prime }=-c$, $b^{\prime }=b$, $c^{\prime }=-a$, then $3\mid c^{\prime }$ and

  $$(a^{\prime }\ne 0)\wedge (b^{\prime }\ne 0)\wedge (c^{\prime }\ne 0).$$

  Then $3\nmid a^{\prime }$ since otherwise by Lemma 3.4 we would have that $3\mid gcd(a,b,c)=1$ which is absurd.
  Analogously, by Lemma 3.3 we have that $3\nmid b^{\prime }$.
  By contradiction we assume that $gcd(a^{\prime },b^{\prime })\ne 1$ which, by basic divisibility properties, implies that there is a prime $p$ such that $p\mid a^{\prime }$ and $p\mid b^{\prime }$. It follows that $p\mid b^{\prime 3}+a^{\prime 3}=b^3-c^3=-a^3$, which implies that $p\mid a$.
  Therefore $p\mid gcd(a,b,c)=1$ which is absurd.
  Moreover, we have that $a^{\prime 3}+b^{\prime 3}=-c^3+b^3=-a^3=c^{\prime 3}$ that contradicts our hypothesis.

• Case $3\mid b$.
  Let $a^{\prime }=a$, $b^{\prime }=-c$, $c^{\prime }=-b$.
  The rest of the proof is analogous to the first case using Lemma 3.3 and Lemma 3.5.

• Case $3\mid c$. Let $a^{\prime }=a$, $b^{\prime }=b$, $c^{\prime }=c$.
  The rest of the proof is analogous to the first case using Lemma 3.4 and Lemma 3.5.

Therefore, we can conclude that $a^3+b^3\ne c^3$.

It directly follows from Lemma 2.10.

Straightforward since $n\in \mathbb{N}$ and $\mathbb{N}$ is well-ordered.

Since $\lambda \nmid a$, then ${\lambda }^4\mid a^3-1\vee {\lambda }^4\mid a^3+1$ by Lemma 2.23. Since $\lambda \nmid b$, then ${\lambda }^4\mid b^3-1\vee {\lambda }^4\mid b^3+1$ by Lemma 2.23. We proceed by analysing each case:

• Case ${\lambda }^4\mid a^3-1\wedge {\lambda }^4\mid b^3-1$. Since $\lambda \mid c$ we have that $\lambda \mid c^3-(a^3-1)-(b^3-1)=2$, which is absurd by Lemma 2.13.

• Case ${\lambda }^4\mid a^3+1\wedge {\lambda }^4\mid b^3+1$. Since $\lambda \mid c$ we have that $\lambda \mid (a^3-1)+(b^3-1)-c^3=2$, which is absurd by Lemma 2.13.

• Case ${\lambda }^4\mid a^3-1\wedge {\lambda }^4\mid b^3+1$. Trivial.

• Case ${\lambda }^4\mid a^3+1\wedge {\lambda }^4\mid b^3-1$. Trivial.

Apply Lemma 3.14 and then compute each case.

Apply Lemma 3.15.

It directly follows from Lemma 3.16.

It directly follows from Lemma 3.17.

Straightforward calculation using Lemma 2.16 and Lemma 2.18.

By contradiction we assume that

Then, by definition, the multiplicity of $\lambda$ in $a+b$, in $a+\eta b$ and in $a+{\eta }^{2}b$ is less than $2$. By properties of divisibility, Lemma 3.16 and Lemma 3.19, we have that

Then, the multiplicity of $\lambda$ in $(a+b)(a+\eta b)(a+{\eta }^{2}b)$ is greater than or equal to $6$.
By Lemma 2.8 $\lambda$ is prime, so we have that the multiplicity of $\lambda$ in $(a+b)(a+\eta b)(a+{\eta }^{2}b)$ is the sum of the multiplicities of $\lambda$ in $a+b$, in $a+\eta b$ and in $a+{\eta }^{2}b$, which is less than $6$. This is a contradiction that forces us to conclude that

By Lemma 3.20, we have that

We proceed by analysing each case:

• Case ${\lambda }^2\mid a+b$. Trivial using $a_1=a$ and $b_1=b$.

• Case ${\lambda }^2\mid a+\eta b$. Let $a_1=a$ and $b_1=\eta b$.
  By Lemma 2.16, we have that $a^3+(\eta b{)}^3=a^3+b^3=u{c}^3$.
  By properties of coprimes and Lemma 2.17, we have that $gcd(a,b)=1$ implies that $gcd(a,\eta b)=1$.
  Since $a_1=a$, we already know that $\lambda \nmid a=a_1$.
  By contradiction we assume that $\lambda \mid b_1=\eta b$, which, by Lemma 2.16, it implies that $\lambda \mid {\eta }^2\eta b=b$ that contradicts our assumption, forcing us to conclude that $\lambda \nmid b_1$.

• Case ${\lambda }^2\mid a+{\eta }^{2}b$. Let $a_1=a$ and $b_1={\eta }^{2}b$.
  By Lemma 2.16, we have that $a^3+({\eta }^{2}b{)}^3=a^3+b^3=u{c}^3$.
  By properties of coprimes and Lemma 2.17, we have that $gcd(a,b)=1$ implies that $gcd(a,{\eta }^{2}b)=1$.
  Since $a_1=a$, we already know that $\lambda \nmid a=a_1$.
  By contradiction we assume that $\lambda \mid b_1={\eta }^{2}b$, which, by Lemma 2.16, it implies that $\lambda \mid \eta {\eta }^{2}b=b$ that contradicts our assumption, forcing us to conclude that $\lambda \nmid b_1$.

Therefore, we can conclude that $\mathrm{\exists }{a}_1,b_1\in {\mathcal{O}}_k$ such that $S_1=(a_1,b_1,c,u)$ is a $solution$.

Let $S^{\prime }=(a^{\prime },b^{\prime },c^{\prime },u^{\prime })$. Let $a,b$ be the units given by Lemma 3.21. Then $S=(a,b,c^{\prime },u^{\prime })$ is a $solutio{n}^{\prime }$ with multiplicity $n$.

Trivial calculation.

Trivial since $\lambda \mid a+b$.

Since $\lambda \mid a+b$, then $\lambda \mid (a+b)+{\lambda }^{2}b+2\lambda b=a+{\eta }^{2}b$.

By contradiction we assume that ${\lambda }^2\mid a+\eta b$, which implies that ${\lambda }^2\mid a+b+\lambda b$ by Lemma 3.23. Since ${\lambda }^2\mid a+b$, then ${\lambda }^2\mid \lambda b$, which implies that $\lambda \mid b$, that contradicts Definition 3.8 forcing us to conclude that ${\lambda }^2\nmid a+\eta b$.

By contradiction using Lemma 2.18, we assume ${\lambda }^2\mid a+{\eta }^{2}b=a+b-b+{\eta }^{2}b$. Since ${\lambda }^2\mid a+b$, then ${\lambda }^2\mid b({\eta }^2-1)=\lambda b(\eta +1)$. Since $\lambda \nmid b$, then $\lambda \mid \eta +1=\lambda +2$, then $\lambda \mid 2$ which is absurd.

Trivial calculation using Lemma 2.18.

We proceed by analysis each case:

• Case $p\mid \lambda$. It directly follows from Lemma 2.8.

• Case $p\nmid \lambda$.
  By hypothesis, we have that $p\mid a+b$ and $p\mid a+\eta b$. Then $p\mid (a+\eta b)-(a+b)=b(\eta -1)=b\lambda$, which implies that $p\mid b$ and we proceed analogously to show that $p\mid a$.
  Therefore $p\mid gcd(a,b)=1$ which is absurd.

Therefore, we can conclude that $p$ is associated with $\lambda$.

We proceed by analysis each case:

• Case $p\mid \lambda$. It directly follows from Lemma 2.8.

• Case $p\nmid \lambda$.
  By hypothesis, we have that $p\mid a+b$ and $p\mid a+{\eta }^{2}b$. By Lemma 2.16 and Lemma 2.17, we have that

  $$p\mid \eta ((a+{\eta }^{2}b)-(a+b))=-({\eta }^3-\eta )b=\lambda b,$$

  which implies that $p\mid b$ and we proceed analogously to show that $p\mid a$.
  Therefore $p\mid gcd(a,b)=1$ which is absurd.

Therefore, we can conclude that $p$ is associated with $\lambda$.

We proceed by analysis each case:

• Case $p\mid \lambda$. It directly follows from Lemma 2.8.

• Case $p\nmid \lambda$.
  By hypothesis, we have that $p\mid a+\eta b$ and $p\mid a+{\eta }^{2}b$. Then $p\mid (a+{\eta }^{2}b)-(a+\eta b)=\eta b(\eta -1)=\eta b\lambda$, which, by Lemma 2.17, implies that $p\mid b$ and we proceed analogously to show that $p\mid a$.
  Therefore $p\mid gcd(a,b)=1$ which is absurd.

Therefore, we can conclude that $p$ is associated with $\lambda$.

By contradiction we assume that $\lambda \mid y$, which implies that ${\lambda }^2\mid \lambda y=a+\eta b$, that contradicts Lemma 3.26 forcing us to conclude that $\lambda \nmid y$.

By contradiction we assume that $\lambda \mid z$, which implies that ${\lambda }^2\mid \lambda z=a+{\eta }^{2}b$, that contradicts Lemma 3.27 forcing us to conclude $\lambda \nmid z$.

By Definition 3.10 we have that ${\lambda }^n\mid c$. Since $u$ is a unit, then by Lemma 3.19 we have that

Then applying Lemma 3.33 and Lemma 3.34, we can conclude that ${\lambda }^{3n-2}\mid a+b$.

By contradiction we assume that $\lambda \mid w$, which implies ${\lambda }^{n+1}\mid {\lambda }^{n}w=c$ that contradicts Definition 3.10 forcing us to conclude that $\lambda \nmid w$.

By contradiction, if $\lambda \mid x$, then ${\lambda }^{3n-1}\mid {\lambda }^{3n-2}x=a+b$. Using Lemma 3.24 and Lemma 3.25, we have that ${\lambda }^{3n+1}\mid (a+b)(a+\eta b)(a+{\eta }^{2}cdotb)=a^3+b^3=u{c}^3=u{\lambda }^{3n}{w}^3$. Then $\lambda \mid w^3$ which implies that $\lambda \mid w$, that contradicts Lemma 3.36 forcing us to conclude $\lambda \nmid x$.

Since $y\ne 0$ by Lemma 3.33, by the properties of PIDs it suffices to prove that $\mathrm{\forall }p\in {\mathcal{O}}_K$ if $p$ is prime and $p\mid x$, then $p\nmid y$. Let $p\in {\mathcal{O}}_K$ be prime and suppose by contradiction that $p\mid x$ and $p\mid y$ which implies that $p\mid {\lambda }^{3n-2}x=a+b$ and $p\mid \lambda y=a+\eta b$. Then by Lemma 3.29 we have that $p$ is associated with $\lambda$, which implies that $\lambda \mid x$ that contradicts Lemma 3.37 forcing us to conclude that $p\nmid y$, which, as stated above, implies that $gcd(x,y)=1$.

Since $z\ne 0$ by Lemma 3.34, by the properties of PIDs it suffices to prove that $\mathrm{\forall }p\in {\mathcal{O}}_K$ if $p$ is prime and $p\mid x$, then $p\nmid z$. Let $p\in {\mathcal{O}}_K$ be prime and suppose by contradiction that $p\mid x$ and $p\mid z$ which implies that $p\mid {\lambda }^{3n-2}x=a+b$ and $p\mid \lambda z=a+{\eta }^{2}b$. Then by Lemma 3.30 we have that $p$ is associated with $\lambda$, which implies that $\lambda \mid x$ that contradicts Lemma 3.37 forcing us to conclude that $p\nmid z$, which, as stated above, implies that $gcd(x,z)=1$.

Since $z\ne 0$ by Lemma 3.34, by the properties of PIDs it suffices to prove that $\mathrm{\forall }p\in {\mathcal{O}}_K$ if $p$ is prime and $p\mid y$, then $p\nmid z$. Let $p\in {\mathcal{O}}_K$ be prime and suppose by contradiction that $p\mid y$ and $p\mid z$ which implies that $p\mid \lambda y=a+\eta b$ and $p\mid \lambda z=a+{\eta }^{2}b$. Then by Lemma 3.31 we have that $p$ is associated with $\lambda$, which implies that $\lambda \mid y$ that contradicts Lemma 3.33 forcing us to conclude that $p\nmid z$, which, as stated above, implies that $gcd(y,z)=1$.

It directly follows from Lemma 3.18 and calculations using ring properties.