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A 3 b 3 c 3 3abc a b ca 2 b 2 c 2 ab bc ca where a 2a b 3b and c 5c now apply values of a b and c on the lhs of identity ie.

A3b3c3 3abc all formula. So either abc0 or a2b2c2 ab ac bc0. Derivation of the formula a3 b3 c3 3abc applications of the formula a3 b3 c3 3abc. A3 b3 c3 3abc a b ca2 b2 c2 ab ac bc.

Therefore abca2b2c2 ab ac bc 0. 2a 3 3b 3 5c 3 32a3b5c and this represents identity. You can re write this as.

A 3 b 3 c 3 3abc a b c a 2 b 2 c 2 ab bc ac if a b c 0 a 3 b 3 c 3 3abc some not so common formulas power n formula a n b n a b a n1 a n2 b 1 a n3 b 2. You didnt specify if abc are all distinct. While considering this question.

Quite unfortunately not from the simple perspective requested i have found that p a 3 b 3 c 3 3abc 2s where s is the area of the triangle of vertices. Given polynomial 8a 3 27b 3 125c 3 90abc can be written as. We know a3 b3 c3 3abc a b c a2 b2 c2 ab ac bc now it is given that.

Value of abc is zero. A 3 b 3 c 3. A 1 b n2 b n1.

Solve 8a 3 27b 3 125c 3 90abc solution.

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$\\ \Rightarrow (a + b + c)^2 = 92\\\\ \text{We have }(a + b + c)^2 =a^2 + b^2 + c^2 + 2(ab + bc + ca) \\\\ \Rightarrow a^2 + b^2 + c^2 + 2(ab + bc + ca) = 81\\\\ \Rightarrow 35 + 2(ab + bc + ca) = 81\\\\ \Rightarrow 2(ab + bc + ca) = 81 - 35 = 46\\\\ \Rightarrow ab + bc + ca = \frac{46}{2}\\\\ \therefore ab + bc + ca = 23$

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$\\Solution:\\\\ Given:\ a+b+c+c=0\\\\\\ \frac{a^2}{bc}+\frac{b^2}{ca}+\frac{c^2}{ab} =\frac{a^3+b^3+c^3}{abc}\\\\\\ We\ know,\\\\ a^3+b^3+c^3=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)+3abc\\\\ \text{If a+b+c=0 then\ } a^3+b^3+c^3=3abc\\\\\\ =\frac{3abc}{abc}\\\\\\ =3$

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