Question

(a+b)³-8(a-b)³
factorise​

Answer 1

ᴛʜᴇ ɢɪᴠᴇɴ ᴀʟɢᴇʙʀᴀɪᴄ ᴇxᴘʀᴇssɪᴏɴ :

(a+b)³-8(a-b)³

➪ (a + b)³ − [2(a-b)] ³

➪ (a + b)³ − [2a − 2b] ³

In order to factorise the given algebraic expression as the differences of two cubes we use the following identities

- a³ - b³ = (a - b) (a² + ab + b²) :

Here a = (a + b) and b = (2a − 2b) :

➪ (a + b − (2a − 2b)) ((a + b)² + (a + b)(2a − 2b) + (2a − 2b)²)

By using an identity (a + b)² = a² + b² + 2ab :

➪ (a + b− 2a + 2b)(a² + b² + 2ab + (a + b)(2a − 2b) + (2a − 2b)²)

On multiplying the terms (a + b)(2a − 2b) :

➪ (a − 2a + b + 2b)(a² + b² + 2ab + 2a² −2ab + 2ab −2b² + (2a − 2b)²)

➪ (3b − a)(3a² + 2ab − b² + (2a − 2b)²)

By using an identity, (a + b)² = a² + b² + 2ab :

➪ (3b − a)(3a² + 2ab − b² + ((2a)² + (2b)² - 2 × 2a × 2b)

➪ (3b − a)(3a² + 2ab − b² + 4a² + 4b² − 8ab)

➪ (3b − a)(3a² + 4a² − b² + 4b² − 8ab + 2ab)

➪ (3b − a)(7a² + 3b² − 6ab)

➪ (- a + 3b)(7a² + 3b² − 6ab)

Hence, the factorization of an algebraic expression is (- a + 3b)(7a² + 3b² − 6ab) .