Factorise:
$8 a^{3} + 125 b^{3} - 64 c^{3} + 120 a b c$

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Solution

We know the identity $a^{3} + b^{3} + c^{3} - 3 a b c = (a + b + c) (a^{2} + b^{2} + c^{2} - a b - b c - c a)$

Using the above identity taking $a = 2 a, b = 5 b$ and $c = - 4 c$ , the equation $8 a^{3} + 125 b^{3} - 64 c^{3} + 120 a b c$ can be factorised as follows:

$8 a^{3} + 125 b^{3} - 64 c^{3} + 120 a b c = (2 a)^{3} + (5 b)^{3} + (- 4 c)^{3} - 3 (2 a) (5 b) (- 4 c)$
$= (2 a + 5 b - 4 c) [(2 a)^{2} + (5 b)^{2} + (- 4 c)^{2} - (2 a \times 5 b) - (5 b \times - 4 c) - (- 4 c \times 2 a)]$
$= (2 a + 5 b - 4 c) (4 a^{2} + 25 b^{2} + 16 c^{2} - 10 a b + 20 b c + 8 c a)$

Hence, $8 a^{3} + 125 b^{3} - 64 c^{3} + 120 a b c = (2 a + 5 b - 4 c) (4 a^{2} + 25 b^{2} + 16 c^{2} - 10 a b + 20 b c + 8 c a)$

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8 a^{3} + 125 b^{3} + 60 a^{2} b + 150 a b^{2}

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

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

3 a + 5 b + 4 c = 0

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

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