x³ + 2y³ + 4z³ = 0

Descent. Suppose a non-zero solution exists, and choose one with $|x|+|y|+|z|$ minimal.

• $x^3=-2y^3-4z^3$ is even, so $x$ is even: $x=2x_1$. Then $8x_1^3+2y^3+4z^3=0$, i.e. $4x_1^3+y^3+2z^3=0$.
• So $y^3$ is even, $y=2y_1$: $4x_1^3+8y_1^3+2z^3=0$, i.e. $2x_1^3+4y_1^3+z^3=0$.
• So $z$ is even, $z=2z_1$: $2x_1^3+4y_1^3+8z_1^3=0$, i.e. $x_1^3+2y_1^3+4z_1^3=0$.

Thus $(x_1,y_1,z_1)=(x/2,y/2,z/2)$ is a smaller non-zero solution, contradiction.

Irrationality. If $\sqrt[3]2=\frac pq$ with $q\ne0$, then $p^3=2q^3$, i.e. $p^3-2q^3=0$. Setting $x=p$, $y=-q$, $z=0$ gives $x^3+2y^3+4z^3=p^3-2q^3=0$ with $(x,y,z)\ne0$, contradiction.

Remark. The same argument proves that $x^3+2y^3+4z^3=0$ has no non-trivial *rational* solutions, which is the statement that $1,\sqrt[3]2,\sqrt[3]4$ are linearly independent over $\mathbb{Q}$.

Lesson. Descent by parity: when all variables are forced to be even, divide by $2$ and obtain a smaller solution.