8. (i) Solve the inequality \( \frac{p+3}{27} \geqslant \frac{1}{2} \). (ii) Find the least possible value of \( p \) such that \( p \) is a perfect cube.

To solve the inequality \( \frac{p+3}{27} \geqslant \frac{1}{2} \), we first multiply both sides by 27 to eliminate the fraction: \( p + 3 \geqslant \frac{27}{2} \). Calculating \( \frac{27}{2} \) gives us 13.5, so we have: \( p + 3 \geqslant 13.5 \). Next, we subtract 3 from both sides: \( p \geqslant 10.5 \). Thus, the solution for \( p \) is \( p \geqslant 10.5 \). For the second part, the least possible integer value of \( p \) that is also a perfect cube and satisfies \( p \geqslant 10.5 \) is \( 3^3 = 27 \), since \( 1^3 = 1, 2^3 = 8 \), and \( 3^3 = 27 \) are the only cubes close to 10.5. So the answer for (i) is \( p \geqslant 10.5 \), and for (ii) it's \( p = 27 \).