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1) Verify the equation $$ \begin{vmatrix} 1 & 1 & 1 \\ a & b & c \\ a^3 &b^3 & c^3 \\ \end{vmatrix}== (a-b)(b-c)(c-a)(a+b+c) $$

2) Show that the points (x1,y1), (x2,y2), (x3,y3) are collinear if

$$ \begin{vmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 &y_3 & 1 \\ \end{vmatrix}=0 $$

For the first problem I get $$ \begin{vmatrix} 1 & 1 & 1 \\ 0 & b-a & c -a\\ 0 &0 & (a^2+ab+b^2)(c-a) - (c^3-a^3) \\ \end{vmatrix} $$

to

$(b-a)(c-a)[a^2 + ab + b^2 - (c^2 - a^2)]$ No idea on how to get the $(a-b)(a+b+c)$

The second problem I have no idea what to do, I'm stuck with

$x_1(y_2-y_3) - y_1(x_2-x_3) + (x_2y_3 - x_3y_2)$

I have heard of something where I must divide numbers and find identical slopes but with no given points I'm practically clueless in this question

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closed as too broad by user147263, Jyrki Lahtonen Oct 7 '15 at 5:35

Please edit the question to limit it to a specific problem with enough detail to identify an adequate answer. Avoid asking multiple distinct questions at once. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

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The transformation to $$ \begin{vmatrix} 1 & 1 & 1 \\ 0 & b-a & c -a\\ 0 &0 & (a^2+ab+b^2)(c-a) - (c^3-a^3) \\ \end{vmatrix} $$ is almost correct, but you must have swapped some row somewhere without changing the sign. Now this one has the determinant $$ (b-a)((a^2+ab+b^2)(c-a) - (c^3-a^3) = (b-a)(c-a)(a^2+ab+b^2-(c^2+ac+a^2)) $$ which simplifies to $$ (a-b)(a-c)(ab+b^2-ac-c^2) = (a-b)(a-c)(b-c)(a+b+c) \, . $$ So your last steps must have some algebraic errors.

For the second problem, collinearity means that there are numbers $a,b,c$, not all zero, such that $ax_i + by_i + c = 0$ for $i = 1,2,3$ (all three points satisfy the same line equation). That is, three separate equations are satisfied by $a,b,c$. Now write these three equations as a matrix vector equation. Then this says that a certain matrix is singular (has a nontrivial null vector) and therefore its determinant is zero. You should be able to take it from there.

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  • $\begingroup$ Thanks for the quick response and help! :) $\endgroup$ – user277665 Oct 7 '15 at 18:37

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