If $2 a + 3 b + c = 0$ then show that:
$8 a^{3} + 27 b^{3} + c^{3} = 18 a b c$

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Solution

We have,

$2 a + 3 b + c = 0$

$2 a + 3 b = - c$ …….. (1)

On taking cube both sides, we get

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

$8 a^{3} + 27 b^{3} + 3 \times 2 a \times 3 b (2 a + 3 b) = - c^{3}$

$8 a^{3} + 27 b^{3} + 18 a b (2 a + 3 b) = - c^{3}$

From equation (1),

$8 a^{3} + 27 b^{3} + 18 a b (- c) = - c^{3}$

$8 a^{3} + 27 b^{3} - 18 a b c = - c^{3}$

$8 a^{3} + 27 b^{3} + c^{3} = 18 a b c$

Hence, proved.

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Similar questions

8 a^{3} + 27 b^{3} + c^{3}

2 a + 3 b + c = 0

If $2 a - 3 b = - 1$ and $a b = 12$

then

$8 a^{3} - 27 b^{3} =$

(2 a + 3 b)

8 a^{3} + 36 a^{2} b + 54 a b^{2} + 27 b^{3}

(3 b - 2 a)

27 b^{3} - 54 b^{2} a + 36 b a^{2} - 8 a^{3}

(2 a - 3 b)^{3}

= 8 a^{3} - 27 b^{3} - 36 a^{2} b + 54 a b^{2}

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