\begin{tabular}{l} (a) Simplify \( \frac{2}{y^{0}} \) \\ (b) Simplify fully \( \left(16 a^{4}\right)^{\frac{3}{4}} \) \\ 5 \\ Factorise fully \( 18 c^{3} d^{2}-21 c^{2} \) \\ (Total for Question 4 is 3 marks) \\ \hline\end{tabular}

To simplify \( \frac{2}{y^{0}} \), remember that any non-zero number raised to the power of 0 equals 1. Thus, \( y^0 = 1 \), so \( \frac{2}{y^{0}} = \frac{2}{1} = 2 \). For \( \left(16 a^{4}\right)^{\frac{3}{4}} \), first, rewrite it using the property of exponents: \( (x^m)^n = x^{m \cdot n} \). Therefore, \( \left(16^{\frac{3}{4}}\right) \cdot \left(a^{4 \cdot \frac{3}{4}}\right) = 16^{\frac{3}{4}} \cdot a^3 \). Since \( 16 = 2^4 \), it becomes \( (2^4)^{\frac{3}{4}} = 2^{3} = 8 \), so the result is \( 8a^{3} \). Next, to factorise \( 18 c^{3} d^{2}-21 c^{2} \), first notice that both terms share common factors. The greatest common factor (GCF) is 3 \( c^{2} \), so factor that out: \( 3 c^{2} (6 cd - 7) \). Voilà! You have the fully factored expression.