Factorise the following?

(i) (a - 1)^3 - 8 + (a + 1)^3 + 6(a^2 - 1)

= (a - 1)^3 + (a + 1)^3 - 8 + 6(a^2 - 1)

First, recognize that a^2 - 1 = a^2 - 1^2 = (a - 1) (a + 1), since it is a difference of two squares. Recall that x^2 - y^2 = (x - y) (x + y)

Also note that (a +1)^3 - 8 = (a + 1)^3 - 2^3 = (a + 1 - 2) [(a + 1)^2 + (a + 1) (2) + 2^2], since it is a difference of two cubes. Recall that x^3 - y^3 = (x - y) [x^2 + xy + y^2]

So we have (a +1)^3 - 8 = (a - 1) [(a + 1)^2 + 2 (a +1) + 4]

Therefore,

(a - 1)^3 - 8 + (a + 1)^3 + 6(a^2 - 1)

= (a - 1)^3 + (a - 1) [(a + 1)^2 + 2 (a +1) + 4] + 6 (a - 1) (a + 1)

Note that (a - 1) is a common factor for all terms, so we have

(a - 1) [(a - 1)^2 + (a + 1)^2 + 2 (a + 1) + 4 + 6(a + 1)]

= (a - 1) [(a^2 - 2a + 1 + a^2 + 2a + 1 + 2a + 2 + 4 + 6a + 6]

= (a - 1) [2a^2 + 8a + 14]

= 2 (a - 1) (a^2 + 4a + 7)

(ii) x^2 - y^2 - 2z^2 + zx + 3yz

= -2z^2 + zx + 3yz + x^2 - y^2

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

= z (x + 3y - 2z) + (x - y) (x + y)

Alternatively,

Note that (x - y + 2z) (x + y - z) = x^2 + xy - xz - xy - y^2 + yz - 2xz + 2yz - 2z^2

= x^2 - y^2 - 2z^2 - 3xz + 3yz

So x^2 - y^2 - 2z^2 + zx + 3yz = (x - y + 2z) (x + y - z) + 4xz

(i)

(a - 1)^3 - 8 + (a + 1)^3 + 6(a² - 1)

= 2 (a - 1) (a^2 + 4 a + 7)

(ii)

x² - y² - 2z² + zx + 3yz

= (x + y - z) (x - y + 2 z)

= (a - 1)³ - 8 + (a + 1)³ + 6.(a² - 1)

= (a - 1)².(a - 1) - 8 + (a + 1)².(a + 1) + 6a² - 6

= (a² - 2a + 1).(a - 1) - 8 + (a² + 2a + 1).(a + 1) + 6a² - 6

= a³ - a² - 2a² + 2a + a - 1 - 8 + a³ + a² + 2a² + 2a + a + 1 + 6a² - 6

= 2a³ + 6a² + 6a - 14

= 2a³ + (8a² - 2a²) + (14a - 8a) - 14

= 2a³ + 8a² - 2a² + 14a - 8a - 14

= 2a³ + 8a² + 14a - 2a² - 8a - 14

= (2a³ + 8a² + 14a) - (2a² + 8a + 14)

= a.(2a² + 8a + 14) - (2a² + 8a + 14)

= (a - 1).(2a² + 8a + 14)

= 2.(a - 1).(a² + 4a + 7)