$\begin{array}{l}a)\,\,\dfrac{1}{8} - y^3\\ = \left(\dfrac{1}{2}\right)^2 - y^3\\ = \left(\dfrac{1}{2} - y\right)\left(\dfrac{1}{4} + \dfrac{1}{2}y + y^2\right)\\ b)\,\,a^3 - (a + b)^3\\ = [a - (a+b)][a^2 + a(a + b) + (a + b)^2]\\ = -b(a^2 + a^2 +ab + a^2 + b^2 + 2ab)\\ = -b(3a^2 + 3ab + b^2)\end{array}$
