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prove that -2a a+b a+c b+a -2b b+c =4(a+b)(b+c)(a+c) a+c b+c -2c

prove that 


-2a  a+b   a+c


b+a  -2b   b+c =4(a+b)(b+c)(a+c)


 


a+c   b+c  -2c

Grade:

1 Answers

Vikas TU
14149 Points
8 years ago
The format is a bit unclear, but it would appear that the determinant is -2a*((-2b)*(-2c) - (b + c)^2) + (a + b)*((b+c)*(a+c) - (b+a)*(-2c)) + (a+c)*((b+a)*(b+c) - (-2b)*(a+c)) = -2a*(4bc - (b^2 + 2bc + c^2)) + (a+b)*(ba + bc + ca + c^2 + 2bc + 2ac) + (a+c)*(b^2 + bc + ab + ac + 2ba + 2bc) = -2a*(2bc - b^2 - c^2) + (a+b)*(ba + 3bc + 3ac + c^2) + (a+c)*(b^2 + 3bc + 3ab + ac) = -4abc + 2a*b^2 + 2a*c^2 + b*a^2 + 3abc + 3c*a^2 + a*c^2 + a*b^2 + 3c*b^2 + 3abc + b*c^2 + a*b^2 + 3abc + 3b*a^2 + c*a^2 + c*b^2 + 3b*c^2 + 3abc + a*c^2 = 8abc + 4a*b^2 +4a*c^2 + 4b*a^2 + 4c*a^2 + 4c*b^2 + 4b*c^2 = 4*(2abc + a*(b^2 + c^2) + b*(a^2 + c^2) + c*(a^2 + b^2)). 4(a+b)(b+c)(a+c) approve plz!

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