Factorise 8 a 3- b 3-4 ax -2 bxA- -2 a-b-4 a2-b2-2 a b-2 x-B- 2 a - b -4 a 2- b - ab -2 x -C- 2 a-b-4 a2-b2-2 a b-2 x-D- 2 a-b-4 a2-b-2 a b-2 x-

The correct option is A =(2a b)[4a^2+b^2+2ab 2x] 8a^3 b^3 4ax+2bx = (2a)^3 b^3 4ax+2bx We know nbsp; a^3 b^3 = (a b)(a^2+ab+b^2) =(2a b)[4a^2+b^2+2ab] 2x(2 ...

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Byju's Answer

Standard IX

Mathematics

Factorization of Expressions Reducible to Difference of Two Squares

Factorise 8 a...

Question

Factorise $8 a^{3} - b^{3} - 4 a x + 2 b x$

= (2 a - b) [4 a^{2} + b^{2} + 2 a b - 2 x]

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(2 a - b) [4 a^{2} + b + a b - 2 x]

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(2 a - b) [4 a^{2} - b^{2} - 2 a b + 2 x]

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(2 a - b) [4 a^{2} + b - 2 a b + 2 x]

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Solution

The correct option is A $= (2 a - b) [4 a^{2} + b^{2} + 2 a b - 2 x]$

$8 a^{3} - b^{3} - 4 a x + 2 b x$
$= (2 a)^{3} - b^{3} - 4 a x + 2 b x$
We know $a^{3} - b^{3} = (a - b) (a^{2} + a b + b^{2})$
$= (2 a - b) [4 a^{2} + b^{2} + 2 a b] - 2 x (2 a - b)$
$= (2 a - b) [4 a^{2} + b^{2} + 2 a b - 2 x]$

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Similar questions

Q. The factors of 8a³ + b³ − 6ab + 1 are

(a) (2a + b − 1) (4a² + b² + 1 − 3ab − 2a)

(b) (2a − b + 1) (4a² + b² − 4ab + 1 − 2a + b)

(c) (2a + b + 1) (4a² + b² + 1 −2ab − b − 2a)

(d) (2a − 1 + b) (4a² + 1 − 4a − b − 2ab)

The factors of $8 a^{3} + b^{3} - 6 a b + 1$ are:

SOLVE:

If 2a+b=11 and ab=24 , find the value of 4a²+b².

$4 a^{2} + b^{2} + 4 a b + 8 a + 4 b + 4$ =?

(a) $(2 a + b + 2)^{2}$

(b) $(2 a - b + 2)^{2}$

(d) None of these

Q. Factorise

4 a^{2} + 4 a b + b^{2}

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Factorise 8a3−b3−4ax+2bx

The correct option is A =(2a−b)[4a2+b2+2ab−2x] 8a3−b3−4ax+2bx =(2a)3−b3−4ax+2bx We know a3−b3=(a−b)(a2+ab+b2) =(2a−b)[4a2+b2+2ab]−2x(2a−b) =(2a−b)[4a2+b2+2ab−2x]

Factorise $8 a^{3} - b^{3} - 4 a x + 2 b x$