JAMB 1989 · UME · Q20

If one factor of $x^3 - 8^{-1}$ is $x - 2^{-1}$ , the other factor is

Worked solution (try it first)

Write the terms as cubes: $8^{-1} = \frac18 = \left(\frac12\right)^3$ , so the expression is $x^3 - \left(\frac12\right)^3$ .
A difference of two cubes factorises as $a^3 - b^3 = (a - b)(a^2 + ab + b^2)$ .
With $a = x$ and $b = \frac12$ : the other factor is $x^2 + \frac12x + \frac14$ , which is $x^2 + 2^{-1}x + 4^{-1}$ , option C.