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in Mathematics by (78 points)
\[ \left|\begin{array}{ccc} b+c & a+b & a \\ c+a & b+c & b \\ a+b & c+a & c \end{array}\right| \] 1) \( a^{3}+b^{3}+c^{3}-3 a b c \) 2) \( 3 a b c-a^{3}-b^{3}-c^{3} \) 3) \( 2\left(a^{3}+b^{5}+c^{3}-3 a b c\right) \) 4) \( 2\left(3 a b c-a^{3}-b^{3}-c^{3}\right. \)

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1 Answer

+1 vote
by (13.0k points)
edited by

\(\left|\begin{array}{ccc} b+c & a+b & a \\ c+a & b+c & b \\ a+b & c+a & c \end{array}\right|\)

\(= \) \((b + c) \begin{bmatrix}b + c&b\\c + a&c\end{bmatrix} - (a + b)\begin{bmatrix}c + a&b\\a + b&c\end{bmatrix} + a\begin{bmatrix}c + a&b + c\\a + b&a + c\end{bmatrix}\)

\( = (b + c)[(bc + c^2) - (bc + ab)] - [(a + b)(c^2 - b^2 + ac - ab)] + a[(a^2 + 2ac + c^2 - (ab + ac + b^2 + bc)]\)

\(= (b + c)(c^2 - ab) - [(a + b)(c^2 - b^2 + ac - ab)] + a(a^2 - b^2 + c^2 + ac - ab - bc)\)

\( = bc^2 - bc^2 - ab^2 - ab^2 + ab^2 + ab^2 - ac^2 + ac^2 - a^2c + a^2c + a^2b - a^2b + a^3 + b^3 + c^3 - 3abc\)

\( = a^3 + b^3 + c^3 - 3abc\)

Option (1) is correct. 

by (78 points)
+2
Thanks a lot
by (13.0k points)
It could have been done in an easy way by applying the properties of determinants but I am not good at that, so my method got longer.

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