Solve: $64 a^{3} - 27 b^{3} - 144 a^{2} b + 108 a b^{2}$ .

(4 a - 9 b)^{3}

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(2 a + 3 b)^{3}

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(4 a - 3 b)^{3}

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(a + 9 b)^{3}

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Solution

The correct option is C $(4 a - 3 b)^{3}$
The given expression $64 a^{3} - 27 b^{3} - 144 a^{2} b + 108 a b^{2}$ can be rewritten as:

$64 a^{3} - 27 b^{3} - 144 a^{2} b + 108 a b^{2} = (4 a)^{3} - (3 b)^{3} - (3 \times (4 a)^{2} \times 3 b) + (3 \times 4 a \times (3 b)^{2}) = (4 a - 3 b)^{3} (∵ (x - y)^{3} = x^{3} - y^{3} - 3 x^{2} y + 3 x y^{2})$

Hence, $64 a^{3} - 27 b^{3} - 144 a^{2} b + 108 a b^{2} = (4 a - 3 b)^{3}$ .

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Similar questions

64 a^{3} - 27 b^{3} - 144 a^{2} b + 108 a b^{2}

= (4 a - 3 b) (4 a - 3 b) (4 a - 3 b)

64 a^{3} - 27 b^{3} - 144 a^{2} b + 108 a b^{2}

64 a^{3}

- 27 b^{3}

- 144 a^{2} b

+ 108 a b^{2}

64 a^{3} - 27 b^{3} - 144 a^{2} b + 108 a b^{2}

Factorise each of the following: 64 $a^{3}$ – 27 $b^{3}$ – 144 $a^{2}$ b + 108a $b^{2}$

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