Prove that \(\frac{sin^3 \ Q + cos^3 \ Q}{sin \ Q + cos \ Q} + \frac{sin^3 \ Q - cos^3 \ Q}{sin \ Q - cos \ Q} = 2\)
Proof:
LHS \(= \frac{sin^3 \ Q + cos^3 \ Q}{sin \ Q + cos \ Q} + \frac{sin^3 \ Q - cos^3 \ Q}{sin \ Q - cos \ Q}\)
By applying the formula \(a^3+b^3=(\)\()(a^2\)\(+b^2)\) ,
\(a^3-b^3=(\)\()(a^2\)\(+b^2)\) then simplifying we get,
\(= 2 \ sin^2 \ Q + 2 \ cos^2 \ Q\)
\(= 2(sin^2 \ Q + cos^2 \ Q)\)
\(= 2(sin^2 \ Q + cos^2 \ Q)\)
By applying the known trigonometric identity, we get:
Hence, proved.