Factorize a^3+b^3+c^3-3abc by factor theorem?
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If you put a = -b-c, you can see that the expression is zero.
(-b-c)^3+b^3+c^3-3(-b-c)bc
=-b^3-3b^2c-3bc^2-c^3 +b^3+c^3+3b^2c+3bc^2 =0
So a+b+c is a factor.
This being cubic adn cyclic in a, b and c, the other factor must be a quadratic and cyclic in a, b and c
So the factors are (a+b+c)(a^2 + b^2 + c^2 - ab - bc - ca)
(-b-c)^3+b^3+c^3-3(-b-c)bc
=-b^3-3b^2c-3bc^2-c^3 +b^3+c^3+3b^2c+3bc^2 =0
So a+b+c is a factor.
This being cubic adn cyclic in a, b and c, the other factor must be a quadratic and cyclic in a, b and c
So the factors are (a+b+c)(a^2 + b^2 + c^2 - ab - bc - ca)
Anonymous:
Awesome ayushi
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