Factorization of 64 a 3-343 b 3 is-A- 4 a-7 b16 a2-49 b2-28 a bB- 4 a-7 b16 a2-49 b2-28 a bC- 4 a-7 b16 a2-49 b2-28 a bD- 4 a-7 b16 a2-49 b2-28 a b

The correct option is A (4a – 7b) (16a2 + 49b2 + 28ab) 64a^3 – 343b^3 = (4a)^3 – (7b)^3 We know that x^3 – y^3 = (x – y)(x^2 + xy + y^2) ∴(4a)^3 – (7b)^3 = ...

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Byju's Answer

Standard IX

Mathematics

(a ± b)²

Factorization...

Question

Factorization of $64 a^{3} - 343 b^{3}$ is:

(4a – 7b) (16a² + 49b² + 28ab)

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(4a + 7b) (16a² - 49b² + 28ab)

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(4a – 7b) (16a² + 49b² - 28ab)

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(4a + 7b) (16a² + 49b² + 28ab)

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Solution

The correct option is A
(4a – 7b) (16a² + 49b² + 28ab)

$64 a^{3} - 343 b^{3} = {(4 a)}^{3} - {(7 b)}^{3}$

We know that $x^{3} - y^{3}$ $=$ $(x - y$ ) $(x^{2} + x y + y^{2}$ )

$∴ {(4 a)}^{3} - {(7 b)}^{3} = (4 a - 7 b) [{(4 a)}^{2} + (7 b) (4 a) + {(7 b)}^{2}]$

$= (4 a - 7 b) (16 a^{2} + 49 b^{2} + 28 a b)$

Suggest Corrections

Factorization of 64a3–343b3 is:

The correct option is A (4a – 7b) (16a2 + 49b2 + 28ab) 64a3–343b3=(4a)3–(7b)3 We know that x3–y3 = (x–y)(x2+xy+y2) ∴(4a)3–(7b)3=(4a–7b)[(4a)2+(7b)(4a)+(7b)2] =(4a–7b)(16a2+49b2+28ab)

Factorization of $64 a^{3} - 343 b^{3}$ is: