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Multiplication of Rational Expressions

Shew that ∑...

Question

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

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Solution

Let $X = a^{2} - b c, Y = b^{2} - a c, Z = c^{2} - a b$
$X + Y + Z = (a + w b + w^{2} c) (a + w^{2} b + w c)$
$X + w Y + w^{2} Z = (a + b + c) (a + w b + w^{2} c)$
$X + w^{2} Y + w Z = (a + b + c) (a + w^{2} b + w c)$
$(X + Y + Z) (X + w Y + w^{2} Z) (X + w^{2} Y + w Z) = \sum X^{3} - 3 X Y Z$
$= {(a + b + c)}^{2} {(a + w b + w^{2} c)}^{2} {(a + w^{2} b + w c)}^{2}$
$= {(a^{3} + b^{3} + c^{3} - 3 a b c)}^{2}$

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\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}

(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}

Q. Prove the following identities:

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

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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}

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