Prove that a2 - b2 - c2 - ab - bc - ca - 1 2- b - c2 - c - a2 - a - b2-

a^2+b^2+c^2 ab bc ac 12[2a^2+2b^2+2c^2 2ab 2bc 2ac] 12[a^2 2ab+b^2+b^2 2bc+c^2+c^2 2ac+a^2] 12[(a b)^2+(b c)^2+(c a)^2] ...

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Types of Expressions Based on the Number of Terms

Prove that ...

Question

Prove that $a^{2} + b^{2} + c^{2} - a b - b c - c a = \frac{1}{2} [{(b - c)}^{2} + {(c - a)}^{2} + {(a - b)}^{2}]$

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a^{2} + b^{2} + c^{2} - a b - b c - c a = 0

a = b = c

a + b + c = 0

\frac{a^{2}}{b c} + \frac{b^{2}}{c a} + \frac{c^{2}}{a b} =

∣ ∣ ∣ ∣ ∣ \begin{matrix} 1 + a^{2} + a^{4} & 1 + a b + a^{2} b^{2} & 1 + a c + a^{2} c^{2} 1 + a b + a^{2} b^{2} & 1 + b^{2} + b^{4} & 1 + b c + b^{2} c^{2} 1 + a c + a^{2} c^{2} & 1 + b c + b^{2} c^{2} & 1 + c^{2} + c^{4} \end{matrix} ∣ ∣ ∣ ∣ ∣ = {(a - b)}^{2} {(b - c)}^{2} {(c - a)}^{2}

If a + b + c = 0, then prove the following

(a) (b + c) (b − c) + a(a + 2b) = 0

(b) a(a² − bc) + b(b² − ca) + c(c² − ab) = 0

(d)

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