State true or false:

If $3 a + 5 b + 4 c = 0$ , then

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

True

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False

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Solution

The correct option is A True
We know that
$(a + b + c)^{3} = (a^{3} + b^{3} + c^{3}) + 3 [(a + b + c) (a b + a c + b c) - a b c]$
If $(a + b + c) = 0$ , then $a^{3} + b^{3} + c^{3} = 3 a b c$
Given, $3 a + 5 b + 4 c = 0$
Here, $a = 3 a$ , $b = 5 b$ and $c = 4 c$
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

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

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

x = 6 a + 8 b + 9 c

y = 2 b - 3 a - 6 c

z = c - b + 3 a

x - y + z

12 a + 5 b + 16 c

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