f.3 maths problems

1.
16(p + q)² - 25q²
= (4p + 4q)² - (5q)²
= [(4p + 4q) + 5q] [(4p + 4q) - 5q]
= (4p + 9q)(4p - q)


2.
9a³b - 12a²b²+ 4ab³
= ab(9a² - 12ab + 4b²)
= ab[(3a)² - 2(3a)(2b) + (2b)²]
= ab(3a - 2b)²


3.
(a + b)³ + (a - b)³
= [(a + b) + (a - b)] [(a + b)² - (a + b)(a - b) + (a - b)²]
= (a + b + a - b)(a² + 2ab + b² - a² + b² + a² - 2ab + b²)
= 2a(a² + 3b²)

2011-09-18 16:51:58 補充：
Identities used :
(u + v)² = u² + 2uv + v²
(u - v)² = u² - 2uv + v²
u² - v² = (u + v)(u - v)
u³ + v³ = (u + v)(u² - uv + v²)

2011-09-18 16:56:31 補充：
Alternative methods:

1.
16(p + q)² - 25q²
= 16p² + 32pq + 16q² - 25q²
= 16p² + 32pq - 9q²
= (4p + 9q)(4p - q)

3.
(a + b)³ + (a - b)³
= a³ + 3a²b + 3ab² + b³ + a³ - 3a²b + 3ab² - b³
= 2a³ + 6ab²
= 2a(a² + 3b²)