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

8 a^{3} + b^{3} + c^{3} - 6 a b c

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8 a^{3} + 2 b^{3} + c^{3} - 8 a b c

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8 a^{3} + b^{3} + 3 c^{3} - 4 a b c

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None of the above

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Solution

The correct option is A $8 a^{3} + b^{3} + c^{3} - 6 a b c$

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

$= (2 a + b + c) [(2 a)^{2} + (b)^{2} + (c)^{2} - (2 a) (b) - (b) (c) - (2 a) (c)]$

$= (x + y + z) (x^{2} + y^{2} + z^{2} - x y - y z - z x = x^{3} + y^{3} + z^{3} - 3 x y z$

Comparing both, we get

$x = 2 a, y = b$ and $z = c$

Then $(2 a + b + c) [(2 a)^{2} + (b)^{2} + (c)^{3} - (2 a) (b) - (b) (c) - (2 a) (c)]$

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

$=$ $8 a^{3} + b^{3} + c^{3} - 6 a b c$

Suggest Corrections

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(2 a + b + c) (4 a^{2} + b^{2} + c^{2} - 2 a b - b c - 2 c a)

8 a^{3} - b^{3} - 4 a x + 2 b x

a + b + c = 0

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

If $a + b + c = 0$ , then the value of $(a + b - c)^{3} + (b + c - a)^{3} + (c + a - b)^{3}$ is:

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