Questions about Spivak's proof that $\sqrt{2} + \sqrt[3]{2}$ is irrational

Let's say that $$x^i = \sum_{j = 0}^5 c_{i,j} n^j. \tag{1}$$

Let $C$ be the matrix $(c_{i,j})$ and let $X = (x^0,\dots,x^6)^T$ and $N = (n^0, \dots, n^5)^T$. Equation $(1)$ says that

$$ X = CN. $$

Now let $A = (a_0, \dots, a_6)$. Then $AX = a_0x^0 + \dots + a_6x^6.$ So we want

$$ AX = ACN = 0. $$

Since $n^0,\dots,n^5$ are linearly independent, this means that $AC = 0$. If you expand this product you get

$$ AC = (b_0, \dots, b_5) = (0,\dots,0) $$

where

$$ b_k = \sum_{j = 0}^6 a_j c_{i,k} = 0. $$

Some additional thoughts:

• Spivak sets $a_6 = 1$ when solving $\sum_{j = 0}^6 a_j c_{i,k} = 0$. This avoids the solution $A = (0,\dots,0)$.

• As Bill Dubuque pointed out in the comments: this works because $\mathbf{Q}[2^{1/6}]$ is a 6-dimensional vector space and $x^0,\dots,x^6$ is seven vectors (necessarily linearly dependent). If you use fewer powers of $x$ you will still get the equation $AC = 0$ but only $A = (0,\dots,0)$ will be a solution.