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Prove That: cos³theta+ sin³theta/cos theta+sintheta

+cos³theta-sin³theta/costheta-sintheta

$\frac{\cos^{3} θ + \sin^{3} θ}{\cos θ + \sin θ} + \frac{\cos^{3} θ - \sin^{3} θ}{\cos θ - \sin θ} = \frac{(\cos θ + \sin θ) (\cos^{2} θ + \sin^{2} θ - \cos θ \sin θ)}{\cos θ + \sin θ} + \frac{(\cos θ - \sin θ) (\cos^{2} θ + \sin^{2} θ + \cos θ \sin θ)}{\cos θ - \sin θ} (u \sin g i d e n t i t y a^{3} + b^{3} = (a + b) (a^{2} + b^{2} - a b) a n d a^{3} - b^{3} = (a - b) (a^{2} + b^{2} + a b)) = \frac{(\cos θ + \sin θ) (1 - \cos θ \sin θ)}{\cos θ + \sin θ} + \frac{(\cos θ - \sin θ) (1 + \cos θ \sin θ)}{\cos θ - \sin θ} = (1 - \cos θ \sin θ) + (1 + \cos θ \sin θ) = 2 H e n c e, \frac{\cos^{3} θ + \sin^{3} θ}{\cos θ + \sin θ} + \frac{\cos^{3} θ - \sin^{3} θ}{\cos θ - \sin θ} = 2$

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