Show that the vectors (1,a,a^2), (1,b,b^2), and (1,c,c^2) are linearly independent if a, b, and c can not equal each other. I am very lost on this one since I am not even sure where to start.
 A: Start with assuming that $\lambda_1 (1, a, a^2)+\lambda_2 (1, b, b^2)+\lambda_3(1,c,c^2)=0$. Then this leads to a system of homogeneous linear equations with coefficient matrix $ \left( \begin{array}{ccc}
1 & 1 & 1 \\
a & b & c \\
a^2 & b^2 & c^2 \end{array} \right)$.
And the determinant of this matrix is $(b-a)(c-b)(c-a)$ which is nonzero. Hence the system only have zero solution.
A: The stated vectors form a Vandermonde Matrix, whose determinant is zero (i.e. linearly dependent rows/columns) if, and only if, either $a=b$, $a=c$ or $b=c$.
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A: If you don't know where to start you should ask yourself what does it mean for some vectors to be linearly independent? A quick Google search will give you definitions and worked out examples. 
Given the vectors $x = \begin{bmatrix}1 \\a \\ a^2 \end{bmatrix}$, $y = \begin{bmatrix}1 \\b \\ b^2 \end{bmatrix}$, $z = \begin{bmatrix}1 \\c \\ c^2 \end{bmatrix}$, you need to show that, given $a\neq b \neq c$ and $r,s,t\in \mathbb{R}$, the equation
$rx + sy + tz = 0 $
can only be satisfied when $r = s= t =0$. That gives you the following three questions:
$r + s + t = 0$, $ra + sb + tc = 0$, and $ra^2 + sb^2 + tc^2 = 0$. Is it possible to solve all three equations with some nonzero $r,s$, or $t$? You need to show that you CAN'T find a non-zero $r,s$ or $t$. You could also show it pairwise: first show $x,y$ are linearly independent, then $x,z$, and then $y,z$. 
