4. Factorise the following \( \begin{array}{l}1.4 .1 x^{3}-8 \\ 1.4 .22 x^{3}+x^{2}-6 x-3 \\ 1.4 .3 x^{4}-16\end{array} \)

To factorise these expressions, let's look at them one by one. 1. \(1.4.1 x^3 - 8\) can be recognised as a difference of cubes, since \(8 = 2^3\). We can rewrite it as \( (x)^3 - (2)^3\), which factors to \( (x - 2)(x^2 + 2x + 4)\). 2. The expression \(1.4.22 x^3 + x^2 - 6x - 3\) can be factored by grouping. Regrouping gives: \( (1.4.22 x^3 + x^2) + (-6x - 3)\). This can be factored as \(x^2(1.4.22 x + 1) - 3(2x + 1)\). This doesn't yield a clean factor, but we can check for rational roots or use synthetic division for further factoring if needed. 3. For \(1.4.3 x^4 - 16\), we can see that this is a difference of squares, noting that \(16 = 4^2\). It factors to \( (x^2 - 4)(x^2 + 4)\), and then \(x^2 - 4\) can be further factored to \( (x - 2)(x + 2)(x^2 + 4)\). Putting it all together, the factorized forms are: 1. \( (x - 2)(x^2 + 2x + 4) \) 2. \( \text{Further factorization needed for } 1.4.22 x^3 + x^2 - 6x - 3 \) 3. \( (x - 2)(x + 2)(x^2 + 4) \)