Find the product $(2 a + 3 b - 4 c) (4 a^{2} + 9 b^{2} + 16 c^{2} - 6 a b + 12 b c + 8 c a)$

8 a^{3} + 27 b^{3} - 64 c^{3} + 72 a b c

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8 a^{3} + 27 b^{3} - 64 c^{3} - 72 a b c

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8 a^{3} - 27 b^{3} - 64 c^{3} + 72 a b c

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none of the above

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Solution

The correct option is A $8 a^{3} + 27 b^{3} - 64 c^{3} + 72 a b c$
We know
$(x + y + z) (x^{2} + y^{2} + z^{2} - x y - y z - z x) = x^{3} + y^{3} + z^{3} - 3 x y z$
$∴ (2 a + 3 b - 4 c) (4 a^{2} + 9 b^{2} + 16 c^{2} + 12 b c + 8 c a)$
$= {(2 a)}^{3} + {(3 b)}^{3} + {(- 4 c)}^{3} - 3 \times (2 a) \times (3 b) \times (- 4 c)$
$= 8 a^{3} + 27 b^{3} - 64 c^{3} + 72 a b c$

Suggest Corrections

Similar questions

{8 a}^{3} - 27 b^{3} - 64 c^{3} - 72 a b c

8 a^{3} + 27 b^{3} + 64 c^{3} - 72 a b c

8 a^{3} + 27 b^{3} + 64 c^{3} - 72 a b c

8a³ + 27b+ 64c³ $-$ 72abc

If $(8 a^{3} + 27 b^{3}) = 0$

then

$(4 a^{2} + 9 b^{2}) =$

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