Show that adding a series capacitor of value C = 1/(w x L) makes the power factor equal 1 in a series RL circuit
Show that the power factor is equal to 1 when C = 1/WL in an RLC
circuit.

Well, you seem to be going down a mathematical route that isn't yielding you the answer you want. Quite simple, for a series tuned circuit at resonance (that also means a power factor of unity): -
$$|X_C| = |X_L|$$
Or, put another way, \$\dfrac{1}{j\omega C} = -j\omega L\$
Therefore, \$1 = \omega^2 LC\hspace{2cm}\$ or \$\hspace{2cm}C = \dfrac{1}{\omega^2 L}\$
So basically your original formula is incorrect because it needs an \$\omega^2\$ term. Or maybe you left that term from your title because in your scribbled notes you appear to say it: -