Shew that - a2 - 2bc3 - 3a2 - 2bc b2 - 2cac2 - 2ab - a3 - b3 - c3 - 3abc2-

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Simplification of an Expression

Shew that ∑...

Question

Shew that $\sum (a^{2} + 2 b c)^{3} - 3 (a^{2} + 2 b c) (b^{2} + 2 c a) (c^{2} + 2 a b) = (a^{3} + b^{3} + c^{3} - 3 a b c)^{2}$ .

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\sum (a^{2} - b c)^{3} - 3 (a^{2} - b c) (b^{2} - c a) (c^{2} - a b) = (a^{3} + b^{3} + c^{3} - 3 a b c)^{2}

(a^{2} + b^{2} + c^{2})^{3} + 2 (b c + c a + a b)^{3} - 3 (a^{2} + b^{2} + c^{2}) (b c + c a + a b)^{2} = (a^{3} + b^{3} + c^{3} - 3 a b c)^{2}

{∣ ∣ ∣ ∣ \begin{matrix} a & b & c b & c & a c & a & b \end{matrix} ∣ ∣ ∣ ∣}^{2} = ∣ ∣ ∣ ∣ ∣ \begin{matrix} 2 b c - a^{2} & c^{2} & b^{2} c^{2} & 2 a c - b^{2} & a^{2} b^{2} & a^{2} & 2 a b - c^{2} \end{matrix} ∣ ∣ ∣ ∣ ∣ = {(a^{3} + b^{3} + c^{3} - 3 a b c)}^{2}

a + b + c + d = 0

\frac{a^{5} + b^{5} + c^{5} + d^{5}}{4} = \frac{a^{3} + b^{3} + c^{3} + d^{3}}{3} \cdot \frac{a^{2} + b^{2} + c^{2} + d^{2}}{2}

(a_{1}, a_{2}, a_{3})

(b_{1}, b_{2}, b_{3})

(c_{1}, c_{2}, c_{3})

x (a_{1}, a_{2}, a_{3}) + y (b_{1}, b_{2}, b_{3}) + z (c_{1}, c_{2}, c_{3}) = 0

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