Question

factorise (2a-b-c)3 +(2b-c-a)3 +(2c-a-b)3

Answer 1

Step-by-step explanation:

a³+b³+c³ = 3abc

(2a-b-c)³+(2b-c-a)³+(2c-a-b)³ = 3(2a-b-c)(2b-c-a)(2c-a-b)

= 6a-3b-3c+6b-3c-3a+6c-3a-3b

Answer 2

Answer:

(2a-b-c)3 +(2b-c-a)3 +(2c-a-b)3 = (2a - b - c)(4a^2 + b^2 + c^2 - 2ab - 2ac + 2bc - 3abc)

Step-by-step explanation:

The given expression can be factorized using the identity for the sum of cubes. The identity states that a^3 + b^3 + c^3 - 3abc = (a + b + c)(a^2 + b^2 + c^2 - ab - ac - bc). Here's how we can use this identity to factorize the expression:

Let's rewrite the given expression as:

(2a)^3 + (-b)^3 + (-c)^3 + 3(2a)(-b)(-c) + 3(-b)(-c)(2b) + 3(-c)(2a)(2b)

We can now apply the identity for the sum of cubes to the first three terms, and we get:

(2a - b - c)(4a^2 + b^2 + c^2 - 2ab - 2ac + 2bc)

The remaining terms can be simplified as:

3(2a)(-b)(-c) + 3(-b)(-c)(2b) + 3(-c)(2a)(2b) = -12abc

Putting it all together, we get:

(2a-b-c)3 +(2b-c-a)3 +(2c-a-b)3 = (2a - b - c)(4a^2 + b^2 + c^2 - 2ab - 2ac + 2bc - 3abc)

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