If $3 a + 5 b + 4 c = 0$ , show that:
$27 a^{3} + 125 b^{3} + 64 c^{3} = 180 a b c$ .

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Solution

We know that
$(a + b + c)^{3} - 3 a b c = (a + b + c) (a^{2} + b^{2} + c^{2} - a b - b c - a c)$
If $(a + b + c) = 0$ , then $a^{3} + b^{3} + c^{3} = 3 a b c$
Given, $3 a + 5 b + 4 c = 0$
Now,
L.H.S $= 27 a^{3} + 125 b^{3} + 64 c^{3}$
$= (3 a)^{3} + (5 b)^{3} + (4 c)^{3}$
$= 3 (3 a) (5 b) (4 c)$
$= 180 a b c$
$=$ R.H.S

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Similar questions

3 a + 5 b + 4 c = 0

27 a^{3} + 125 b^{3} + 64 c^{3} = 180 a b c

If 3a + 5b + 4c = 0, Find the value of $27 a^{3} + 125 b^{3} + 64 c^{3}$ .

27 a^{3} - (3 a - b)^{3}

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