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Show that $a+b+c$ is a factor of $$\det \begin{pmatrix} b+c & a & a^3\\ c+a & b & b^3\\ a+b & c & c^3 \end{pmatrix}$$ and express the determinant as a product of five factors.

I was able to get the first phase of the question but the second part proved difficult for me. The answer is $$(a+b+c)^2(a-b)(b-c)(c-a)$$ but how to get it remains a mystery to me.

Similar problem like this was given me. Here I just want to be sure if am correct. Express $\det \begin{bmatrix}1 & 1 & 1\\ a^2 & b^2 & c^2\\ [a+b]^2 & [c+a]^2 & [a+b]^2\end{bmatrix}$ as a product of linear factors After solving on my own and following the guide provided here. I got $(b-a)(b+a)(b-c)(2a+b+c)$. However, I was meant to understand that the correct solution to the question is $2(a-b)(b-c)(a+b+c)$ So which is the correct answer? Mine or the latter?

Michael Umande
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  • I just can identify 8 values only at {{b+c} a b^3 {c+a} b^3 {a+b} c c^3}. Please make a check. – callculus42 Oct 17 '18 at 11:52
  • Okay this is a 3 by 3 determinant. I just don't know how to write it using mathjax. However, let's say A=Det. A_11=b+c A_12=a A_13=a³ A_21=c+a A_22=b A_23=b³ A_31 a+b A_32=c A_33=c³ – Michael Umande Oct 17 '18 at 11:57
  • Is this your determinant? $\det \begin{bmatrix} b+c & a & b^3\c+a & b& b^3\a+b & c & c^3 \end{bmatrix}$ – user376343 Oct 17 '18 at 12:09
  • Exactly. To express the determinant as a product of five factors – Michael Umande Oct 17 '18 at 12:11
  • @user376343 It looks like the $1,3$ entry should be $a^3$, not $b^3$. – paw88789 Oct 17 '18 at 12:13
  • Yeah it's $a^3$ and not $b^3$. – Michael Umande Oct 17 '18 at 12:16
  • Click on my determinant with the right mouse button. In "Show Math As" choose "TeX Commands". Then copy it and paste in your question (you have to select "edit"). Dont forget to embrace each formula by dollars $ (one at the beginning, one at the end of the formula) Then try to edit your question so we can understand. – user376343 Oct 17 '18 at 12:23
  • I've done so. Any attempt to the question will be really appreciated – Michael Umande Oct 17 '18 at 12:53

1 Answers1

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$$\begin{array}{ccc} \det\left(\begin{array}{ccc}b+c&a&a^3\\c+a&b&b^3\\ a+b&c&c^3\\ \end{array} \right) &\stackrel{c_2+c_1\to c_1}{=} & \det\left(\begin{array}{ccc}a+b+c&a&a^3\\a+b+c&b&b^3\\ a+b+c&c&c^3& \end{array} \right)\\ \\ &=& (a+b+c)\cdot \det\left(\begin{array}{ccc}1&a&a^3\\1&b&b^3\\ 1&c&c^3& \end{array} \right)\\ \\ &\stackrel{-r_2+r_1 \to r_1}{=}& (a+b+c)\cdot \det\left(\begin{array}{ccc}0&a-b&a^3-b^3\\1&b&b^3\\ 1&c&c^3& \end{array} \right)\\ \\ &=&(a+b+c)\cdot(a-b)\cdot \det\left(\begin{array}{ccc}0&1&a^2+ab+b^2\\1&b&b^3\\ 1&c&c^3& \end{array} \right)\\ \\ &\stackrel{-r_3+r_2\to r_2}{=}& (a+b+c)\cdot (a-b)\cdot \det\left(\begin{array}{ccc}0&1&a^2+ab+b^2\\0&b-c&b^3-c^3\\ 1&c&c^3& \end{array} \right)\\ \\ &=& (a+b+c)\cdot(a-b)\cdot(b-c)\cdot \det\left(\begin{array}{ccc}0&1&a^2+ab+b^2\\0&1&b^2+bc+c^2\\ 1&c&c^3& \end{array} \right)\\ \\ &=& (a+b+c)\cdot(a-b)\cdot(b-c)\cdot((b^2+bc+c^2)-(a^2+ab+b^2)) \\ \\ &=& (a+b+c)\cdot(a-b)\cdot(b-c)\cdot (bc+c^2-a^2-ab) \\ \\ &=& (a+b+c)\cdot(a-b)\cdot(b-c)\cdot ((c-a)(c+a)+b(c-a))\\ \\ &=& (a+b+c)\cdot(a-b)\cdot(b-c)\cdot ((c-a)(c+a+b)) \\ \\ &=& (a+b+c)^2\cdot(a-b)\cdot(b-c)\cdot(c-a) \end{array}$$

paw88789
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