Question

Factorise:
(a - 2b)^3 + (2a - b)^3

Answer 1

Explanation:

To factorise the expression, we can use the factor theorem which states that if a polynomial p(x) has a factor (x - r), then p(r) = 0.

We can use this theorem to find the factors of the polynomial expression:

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

We can write (a - 2b)^3 as (a - 2b)(a^2 - 4ab + 4b^2). Using the factor theorem, we see that:

(a - 2b)(a^2 - 4ab + 4b^2) = 0 when a = 2b

Similarly, we can write (2a - b)^3 as (2a - b)(4a^2 - 4ab + b^2). Using the factor theorem, we see that:

(2a - b)(4a^2 - 4ab + b^2) = 0 when a = b/2

Now we have two equations: a = 2b and a = b/2, which we can use to solve for a and b in terms of each other. We see that:

a = 2b = b/2

Multiplying both sides by 2, we get:

2a = b

We can substitute this back into the original expression:

(a - 2b)^3 + (2a - b)^3 = (a - 2(2a))^3 + (2a - (2a))^3 = (-3a)^3 + (0)^3 = -27a^3

So the factorised form of the expression is:

(a - 2b)^3 + (2a - b)^3 = -27a^3.

Answer 2

Answer: 9(a-b)(a^2-ab+b^2)

Work Shown:

Use the sum of cubes factoring rule

x^3 + y^3 = (x+y)(x^2 - xy + y^2)

In this case,

• x = a-2b
• y = 2a-b

So,

x^3 + y^3 = (x+y)(x^2 - xy + y^2)

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

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

(a-2b)^3 + (2a-b)^3 = (3a-3b)((a^2-4ab+4b^2) - (2a^2-5ab+2b^2) + (4a^2-4ab+b^2))

(a-2b)^3 + (2a-b)^3 = (3a-3b)(a^2-4ab+4b^2 - 2a^2+5ab-2b^2 + 4a^2-4ab+b^2)

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

(a-2b)^3 + (2a-b)^3 = 3*3(a-b)(a^2-ab+b^2)

(a-2b)^3 + (2a-b)^3 = 9(a-b)(a^2-ab+b^2)

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