Find the product
$(2 a + b + c) (4 a^{2} + b^{2} + c^{2} - 2 a b - b c - 2 c a)$

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Solution

By using the identity
$a^{3} + b^{3} + c^{3} - 3 a b c = (a + b + c) (a^{2} + b^{2} + c^{2} - a b - b c - c a)$
$∴ (2 a + b + c) (4 a^{2} + b^{2} + c^{2} - 2 a b - b c - 2 c a)$
$= 8 a^{3} + b^{3} + c^{3} - 4 a^{2} - 2 a b - 2 a b^{2} - 2 a b c - b^{2} c - b c^{2} - 2 a^{2} c - 2 a b c - 2 a c^{2} + 4 a^{2} b + 4 a^{2} c + 2 a b^{2} + b^{2} c + 2 a c^{2} + b c^{2}$
$= 8 a^{3} + b^{3} + c^{3} - 6 a b c$

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Similar questions

If a + b + c = 2s, then prove the following identities

(a) s² + (s − a)² + (s − b)² + (s − c)² = a² + b² + c²

(b) a² + b² − c² + 2ab = 4s (s − c)

(d) a² − b² − c² + 2ab = 4(s − b) (s − c)

(e) (2bc + a² − b² − c²) (2bc − a² + b² + c²) = 16s (s − a) (s − b) (s − c)

(f)

(a - b + c)^{2}

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

△ A B C

cos A = \frac{b^{2} + c^{2} - a^{2}}{2 b c}

cos B = \frac{c^{2} + a^{2} - b^{2}}{2 c a}, cos C = \frac{a^{2} + b^{2} - c^{2}}{2 a b}

(a + b - c)^{2}

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

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