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Prove without expanding: \begin{equation}\begin{vmatrix}1&1&1\\a^2&b^2&c^2\\a^3&b^3 & c^3\end{vmatrix} = (ab + ac + bc)(b - a)(c - a)(c - b)\end{equation}


  • I tried to zero some elements and expand until I reach the Right hand side.
  • Also tried C1-C3, C2-C3 then decompose the determinant into two determinants and taking common factors. But I couldn't get (ab + ac + bc) part.
  • I can only use the properties shown here http://www.vitutor.com/alg/determinants/properties_determinants.html
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    • $\begingroup$ What have you tried? It looks like cofactor expansion across the top row would be a good idea. $\endgroup$ – felani May 8 '15 at 15:17
    • $\begingroup$ The minor associatedd to the upper left hand entry is $b^{2}c^{3}-c^{2}b^{3}$. Include the other two minors, expand and factorise. $\endgroup$ – Autolatry May 8 '15 at 15:19
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      $\begingroup$ You're telling us what we should do? $\endgroup$ – GFauxPas May 8 '15 at 15:21
    • $\begingroup$ @GFauxPas Aww, there's no need to be like that. $\endgroup$ – Kitegi May 8 '15 at 15:32
    • $\begingroup$ Still unclear. The recent Edit adjures Readers to prove "without expanding by using determinant properties". However the purpose of doing A without B is not explained, nor is it clear what remains allowed if using determinant properties is forbidden. $\endgroup$ – hardmath May 12 '15 at 22:23
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    First note that the determinant is cyclic. Hence, it is of the form $f(a,b,c)$, where $f$ is a polynomial of degree $5$. Further, we have $f(a,a,c) = f(a,b,b) = f(c,b,c) = 0$, which means $(a-b)$, $(b-c)$ and $(c-a)$ are factors, i.e., the determinant is $g(a,b,c)(a-b)(b-c)(c-a)$, where $g(a,b,c)$ is a cyclic polynomial of degree $2$. Any cyclic polynomial of degree $2$ is of the form $x(a^2+b^2+c^2)+y(ab+bc+ca)$. Setting $a=0$, we see that the determinant is $b^2c^3-b^3c^2 = b^2c^2(c-b)$. We also have $$f(0,b,c) = g(0,b,c)(-b)(b-c)c = \left(x(b^2+c^2)+ybc \right)bc(c-b)$$ This means we need $$\left(x(b^2+c^2)+ybc \right)bc(c-b) = b^2c^2(c-b) \implies x(b^2+c^2)+ybc = bc \implies x=0,y=1$$ Hence, we obtain that the determinant is $$(ab+bc+ca)(a-b)(b-c)(c-a)$$

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    \begin{gather*} \begin{bmatrix}1&1&1\\a^2&b^2&c^2\\a^3&b^3&c^3\end{bmatrix} \\ = \begin{bmatrix}1&0&0\\a^2&b^2-a^2&c^2-a^2\\a^3&b^3-a^3&c^3 - a^3\end{bmatrix} \\ = \begin{bmatrix}b^2-a^2&c^2-a^2\\b^3-a^3&c^3 - a^3\end{bmatrix} \\ = (b^2-a^2)(c^3 - a^3)-(b^3-a^3)(c^2 - a^2) = (b-a)(c-a)[(b+a)(c^2+ac+a^2)-(c+a)(b^2+ab+a^2)] \\ = (ab+ac+bc)(b-a)(c-a)(c-b)\\ = R.H.S \end{gather*}

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    I found a long proof but that's the only way I could answer it. \begin{gather*} \begin{bmatrix}1&1&1\\a^2&b^2&c^2\\a^3&b^3&c^3\end{bmatrix} \\ = \begin{bmatrix}1&0&0\\a^2&b^2-a^2&c^2-a^2\\a^3&b^3-a^3&c^3 - a^3\end{bmatrix} \\ = (b-a)(c-a)\begin{bmatrix}1&0&0\\a^2&b+a&c+a\\a^3&b^2+ab+a^2&c^2+ac+a^2\end{bmatrix} \\ = (b-a)(c-a)\begin{bmatrix}1&0&0\\a^2&b+a&c-b\\a^3&b^2+ab+a^2&c^2-b^2+ac-ab\end{bmatrix} \\ = (b-a)(c-a)(c-b)\begin{bmatrix}1&0&0\\a^2&b+a&1\\a^3&b^2+ab+a^2&a+b+c\end{bmatrix}\\ = (b-a)(c-a)(c-b)[(b+a)(a+b+c)-(b^2+ab+a^2)] \\ = (ab+ac+bc)(b-a)(c-a)(c-b)\\ = R.H.S \end{gather*}

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