Set $\displaystyle \alpha = \sqrt{2}$ and $\displaystyle \beta= \sqrt{3} $. Note that $\displaystyle \alpha^2 = 2$ and $\displaystyle \beta^2 = 3$. $\displaystyle Q(\alpha, \beta) $ is the 4 dimensional vector space of R having basis $\displaystyle \{1, \alpha, \beta, \alpha \beta \}$.

**(a)** Show that real numbers $\displaystyle (\alpha, \beta)^2$ and $\displaystyle (\alpha, \beta)^4$ are both in space $\displaystyle Q(\alpha, \beta) $ by expanding the powers and expressing them as a linear combination of the basis vectors.

**(b)** Show that the vectors $\displaystyle \{(\alpha + \beta)^4), (\alpha+ \beta)^2,1\}$ are linearly dependent.

My Attempt:

**(a)** $\displaystyle (\alpha, \beta)^2 = \alpha^2 + 2 \alpha \beta + \beta^2 = 2+ 2 \alpha \beta +3$, which can be written as:

$\displaystyle 2.1 + 2 \alpha \beta +3 . 1$ (since $\displaystyle 1 \in Q(\alpha, \beta)$ and $\displaystyle 2,3 \in R$)

Now, $\displaystyle (\alpha, \beta)^4 = \alpha^4 + 4 \alpha^3 \beta + 6 \alpha^2 \beta^2 + 4 \alpha \beta^3 + b^4$

$\displaystyle (2.2).1+(4.2) \alpha \beta + (6.2.3).1+(3.4)\alpha \beta + (4).1$ $\displaystyle = (4).1+(8) \alpha \beta + (46).1+(12) \alpha \beta + (4).1$

Is this right?

**(b)** This is the HARD part! I know that the vectors $\displaystyle \{(\alpha + \beta)^4), (\alpha+ \beta)^2,1\}$ are linearly dependent if there are $\displaystyle \lambda_1, \lambda_2, \lambda_3$ in reals, NOt all 0, such that

$\displaystyle \lambda_1 (\alpha + \beta)^4), \lambda_2 (\alpha+ \beta)^2,\lambda_3 1 = 0$

But I cant' figure out how to prove/represent this. Can anyone show me?