This is generally seen as a polynomial question and in terms of alpha beta and gamma, rather than a, b and c but my book doesn't explain it AT ALL it just says its too complicated and just gives a simple answer (which i don't like as knowing how something is done is much better than just knowing what it is). Thanks and goodluck (the question arises in Ext. 2 maths [NSW] and also deviates to the 4th degree)
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1) Consider a^3 + b^3 + c^3 - 3abc
2) If c = -(a+b), then,
a^3 + b^3 + c^3 - 3abc = a^3 + b^3 - (a+b)^3 - 3ab*{-(a+b)} = {a^3 + b^3 + 3ab(a+b)} - (a+b)^3 = 0
Hence by factor theorem, (a+b+c) is a factor of a^3 + b^3 + c^3 - 3abc
3) Thus a^3 + b^3 + c^3 - 3abc ≅ (a+b+c)*{p(a^2 + b^2 + c^2) + q(ab + bc + ca)}
[Note: As left side is a homogeneous expression of 3rd degree polynomial, when it is factored the right side also must be a third degree; to balance this, we are taking the other factor as, {p(a^2 + b^2 + c^2) + q(ab + bc + ca)}].
4) Since this being a congruence, we can plug some real values for a, b & c on both sides and evaluate for 'p' and 'q'.
So, plugging (a,b,c) = (0,1,2) and (1,2,3), we get two equations as:
5p + 2q = 3 and 14p + 11q = 3
Solving these two equations, we get, p =1 and q = -1.
5) Thus Thus a^3 + b^3 + c^3 - 3abc = (a+b+c)*{(a^2 + b^2 + c^2) - (ab + bc + ca)}
Hence a^3 + b^3 + c^3 = (a+b+c)*(a^2 + b^2 + c^2 - ab - bc - ca) + 3abc
2) If c = -(a+b), then,
a^3 + b^3 + c^3 - 3abc = a^3 + b^3 - (a+b)^3 - 3ab*{-(a+b)} = {a^3 + b^3 + 3ab(a+b)} - (a+b)^3 = 0
Hence by factor theorem, (a+b+c) is a factor of a^3 + b^3 + c^3 - 3abc
3) Thus a^3 + b^3 + c^3 - 3abc ≅ (a+b+c)*{p(a^2 + b^2 + c^2) + q(ab + bc + ca)}
[Note: As left side is a homogeneous expression of 3rd degree polynomial, when it is factored the right side also must be a third degree; to balance this, we are taking the other factor as, {p(a^2 + b^2 + c^2) + q(ab + bc + ca)}].
4) Since this being a congruence, we can plug some real values for a, b & c on both sides and evaluate for 'p' and 'q'.
So, plugging (a,b,c) = (0,1,2) and (1,2,3), we get two equations as:
5p + 2q = 3 and 14p + 11q = 3
Solving these two equations, we get, p =1 and q = -1.
5) Thus Thus a^3 + b^3 + c^3 - 3abc = (a+b+c)*{(a^2 + b^2 + c^2) - (ab + bc + ca)}
Hence a^3 + b^3 + c^3 = (a+b+c)*(a^2 + b^2 + c^2 - ab - bc - ca) + 3abc
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Everyone is looking for the factored version, but a factored version isn't the simplified version. In fact, that is already simplified.
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(a+b+c)^3 = a^3 + (b+c)^3 + 3 a^2 (b+c) + 3 a(b+c)^2
(a+b+c)^3 = a^3 + b^3 c^3 + 3b^2 c + 3bc^2 + 3a^2b + 3a^2 c + 3ab^2 + 3ac^2 + 6abc
(a+b+c)^3 - 3b^2 c - 3bc^2 - 3a^2b - 3a^2 c - 3ab^2 - 3ac^2 - 6abc = a^3 + b^3 + c^3
(a+b+c)^3 = a^3 + b^3 c^3 + 3b^2 c + 3bc^2 + 3a^2b + 3a^2 c + 3ab^2 + 3ac^2 + 6abc
(a+b+c)^3 - 3b^2 c - 3bc^2 - 3a^2b - 3a^2 c - 3ab^2 - 3ac^2 - 6abc = a^3 + b^3 + c^3
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[(a+b) (a^2-ab+b^2)] + c^3
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(a+b+c)^3 I suppose
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you cant