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(i) Factorize the expression 8 a⁶ + 5 a³ + 1. (ii) Prove the identity (x - y)³ + (y - z)³ + (z - x)³ = 3(x - y)(y - z)(z - x).

1. (i) does not factor over rationals; (ii) identity holds since the three cubed terms sum to zero
2. (i) (2a³ + 1)(4a³ + 1); (ii) holds
3. (i) (8a³ + 1)(a³ + 5); (ii) holds
4. (i) (2a² + 1)(4a⁴ + 1); (ii) holds

Correct answer: (i) does not factor over rationals; (ii) identity holds since the three cubed terms sum to zero

Solution

(i) Writing t = a³ gives 8t² + 5t + 1, whose discriminant 5² - 4*8*1 = 25 - 32 = -7 is negative, so it has no real (hence no rational) linear factors; it does not factor nicely over the rationals. (ii) Let p = x - y, q = y - z, r = z - x. Then p + q + r = 0, and the identity p³ + q³ + r³ = 3 p q r gives the result directly.

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