[tex]\displaystyle\bf\\\Big(\frac{1}{3} c-\frac{3}{13}d\Big) \cdot\Big(\frac{1}{3} c+\frac{3}{13}d\Big) =\Big(\frac{1}{3} c\Big)^{2} -\Big(\frac{3}{13}d\Big)^{2} =\frac{1}{9}c^{2}-\frac{9}{169} d^{2}[/tex]