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Without using the Rule of Sarrus, prove that: $$\left| \begin{matrix} (b+c)&(a-b)&a \\ (c+a)&(b-c)&b \\ (a+b)&(c-a)&c \\ \end{matrix}\right|=3abc-a^3-b^3-c^3$$

My Approach: $$LHS= \left| \begin{matrix} (b+c)&(a-b)&a \\ (c+a)&(b-c)&b \\ (a+b)&(c-a)&c \\ \end{matrix}\right|$$ $$C_1\to C_1+C_2$$ $$= \left| \begin{matrix} (c+a)&(a-b)&a \\ (a+b)&(b-c)&b \\ (b+c)&(c-a)&c \\ \end{matrix}\right|$$ $$C_1\to C_1-C_3$$ $$= \left| \begin{matrix} c&(a-b)&a \\ a&(b-c)&b \\ b&(c-a)&c \\ \end{matrix}\right|$$

How do I complete the rest?

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  • $\begingroup$ pick a row or column and use the determinant formula and the answer should come immediately. $\endgroup$ Apr 27, 2018 at 13:33
  • $\begingroup$ @JohnColtraneis, That would be a quick and direct expansion which I don't want to use. I'm wanting for further simplification before expansion so that there may be some 0's in any row or column. $\endgroup$
    – pi-π
    Apr 27, 2018 at 13:36
  • $\begingroup$ Sorry I just saw your comment before I added my answer $\endgroup$ Apr 27, 2018 at 13:45
  • $\begingroup$ Obviously with your reputation you knew that, I don't know what I was thinking, my mistake. $\endgroup$ Apr 27, 2018 at 13:46

4 Answers 4

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Here is a way to break it down to a factor and only one $2\times 2$-determinant containing only binomials before expanding: $$\left| \begin{matrix} (b+c)&(a-b)&a \\ (c+a)&(b-c)&b \\ (a+b)&(c-a)&c \\ \end{matrix}\right|\stackrel{R_3 \mapsto R_3+R_2+R_1}{=} \left| \begin{matrix} (b+c)&(a-b)&a \\ (c+a)&(b-c)&b \\ 2(a+b+c)&0&(a+b+c) \\ \end{matrix}\right| = (a+b+c)\left| \begin{matrix} (b+c)&(a-b)&a \\ (c+a)&(b-c)&b \\ 2&0&1 \\ \end{matrix}\right| \stackrel{C_1 \mapsto C_1-2C_3}{=} (a+b+c)\left| \begin{matrix} (b+c-2a)&(a-b)&a \\ (c+a-2b)&(b-c)&b \\ 0&0&1 \\ \end{matrix}\right| \stackrel{C_1 \mapsto C_1+C_2}{=} (a+b+c)\left| \begin{matrix} (c-a)&(a-b)\\ (a-b)&(b-c) \end{matrix}\right| =$$ $$= (a+b+c)(ab+ac+bc-a^2-b^2-c^2) = 3abc - (a^3 + b^3 + c^3) $$

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You're close. Perform $C_2 \mapsto C_2 - C_3$. Then you have

$$- \begin{vmatrix} c & b & a \\ a & c & b \\ b & a & c\end{vmatrix}$$

Now apply the determinant formula, $$-(c(c^2 - ba) - b(ac - b^2) + a(a^2 - cb)) = 3abc - a^3 - b^3 - c^3$$

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    $\begingroup$ The point was to avoid using the determinant formula, wasn't it? $\endgroup$ Apr 27, 2018 at 14:09
  • $\begingroup$ It says "without using the Rule of Sarrus," not the Leibniz formula. Admittedly, Sarrus is much cleaner. However, I think the point was to simplify until we reached a nice form to apply the formula. I don't think there is a nice geometric interpretation with wedges, if there is it is probably too complex for what was intended. $\endgroup$
    – user211599
    Apr 27, 2018 at 14:15
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\begin{align*} \left| \begin{matrix} c&(a-b)&a \\ a&(b-c)&b \\ b&(c-a)&c \\ \end{matrix}\right|&= c\left|\begin{matrix} (b-c)&b \\ (c-a)&c \\ \end{matrix}\right|-a\left|\begin{matrix}(a-b)&a\\(c-a)&c \\\end{matrix}\right|+b\left|\begin{matrix}(a-b)&a\\(b-c)&b \\\end{matrix}\right|\\ &=c[bc-c^2-bc+ab]-a[ac-bc-ac+a^2]+b[ab-b^2-ab+ac]\\ &=-c^3+abc+abc-a^3-b^3+abc\\ &=3abc-a^3-b^3-c^3. \end{align*}

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\begin{align} \left| \begin{matrix} (b+c)&(a-b)&a \\ (c+a)&(b-c)&b \\ (a+b)&(c-a)&c \\ \end{matrix}\right|&= \left| \begin{matrix} a+b+c&a-b&a \\ a+b+c&b-c&b \\ a+b+c&c-a&c \\ \end{matrix}\right|\\ &= (a+b+c)\left| \begin{matrix} 1&a-b&a \\ 1&b-c&b \\ 1&c-a&c \\ \end{matrix}\right|\\ &= (a+b+c)\left| \begin{matrix} 1&-b&a \\ 1&-c&b \\ 1&-a&c \\ \end{matrix}\right|\end{align} Can you continue from here?

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