Question

Solve: (3a²b/2a²c)x(3a²/2b²)÷(27a²/8b²)²​

Answer 1

The factorise the given expression a^2-4b^2+a^3-8b^3-(a-2b)^2 is (a-2b)(4b+a^{2}+2ab+4b^{2}).

Step-by-step explanation:

We have,

a^2-4b^2+a^3-8b^3-(a-2b)^2

To factorise the given expression=?

=(a^2-4b^2)+(a^3-8b^3)-(a-2b)^2

=[a^2-(2b)^2]+[a^3-(2b)^3]-(a-2b)^2

=(a+2b)(a-2b)+(a-2b)(a^{2}+2ab+4b^{2})-(a-2b)^2

Using identities,

a^{2}-b^{2} =(a+b)(a-b) and

a^{3}-b^{3} =(a-b)(a^{2}+ab+b^{2})

Taking (a-2b) as common, we get

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

=(a-2b)(a+2b+a^{2}+2ab+4b^{2}-a+2b)

=(a-2b)(4b+a^{2}+2ab+4b^{2})

=(a-2b)(4b+a^{2}+2ab+4b^{2})

Hence, the factorise the given expression a^2-4b^2+a^3-8b^3-(a-2b)^2 is (a-2b)(4b+a^{2}+2ab+4b^{2}).

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Answer 2

The factorise the given expression a^2-4b^2+a^3-8b^3-(a-2b)^2 is (a-2b)(4b+a^{2}+2ab+4b^{2}).

Step-by-step explanation:

We have,

a^2-4b^2+a^3-8b^3-(a-2b)^2

To factorise the given expression=?

=(a^2-4b^2)+(a^3-8b^3)-(a-2b)^2

=[a^2-(2b)^2]+[a^3-(2b)^3]-(a-2b)^2

=(a+2b)(a-2b)+(a-2b)(a^{2}+2ab+4b^{2})-(a-2b)^2

Using identities,

a^{2}-b^{2} =(a+b)(a-b) and

a^{3}-b^{3} =(a-b)(a^{2}+ab+b^{2})

Taking (a-2b) as common, we get

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

=(a-2b)(a+2b+a^{2}+2ab+4b^{2}-a+2b)

=(a-2b)(4b+a^{2}+2ab+4b^{2})

=(a-2b)(4b+a^{2}+2ab+4b^{2})

Hence, the factorise the given expression a^2-4b^2+a^3-8b^3-(a-2b)^2 is (a-2b)(4b+a^{2}+2ab+4b^{2}).

HOPE MY ANSWER WILL BE HELPFUL FOR YOU DEAR :)

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