(i) Start with \(x_1 = 3\).

Calculate \(x_2 = \frac{3 \times 3}{4} + \frac{15}{3^3} = 2.75\).

Calculate \(x_3 = \frac{3 \times 2.75}{4} + \frac{15}{2.75^3} = 2.7816\).

Calculate \(x_4 = \frac{3 \times 2.7816}{4} + \frac{15}{2.7816^3} = 2.7794\).

Calculate \(x_5 = \frac{3 \times 2.7794}{4} + \frac{15}{2.7794^3} = 2.7800\).

Calculate \(x_6 = \frac{3 \times 2.7800}{4} + \frac{15}{2.7800^3} = 2.7798\).

Since the values are converging, \(\alpha \approx 2.78\) to 2 decimal places.

(ii) The equation satisfied by \(\alpha\) is \(x = \frac{3}{4}x + \frac{15}{x^3}\).

Rearrange to find \(x^4 = 60\), so \(\alpha = \sqrt[4]{60}\).