Homework help with trigonometry? please?

Question

simplify the expression: (sin^(3)θ+cos^(3)θ)/(sinθ+cosθ)
i know the answer from the back of the book but i cant figure out how to get it.
:'''( iv been stuck on this for hours and cant figure it out.
thanks :(

Answer 1

simplify the expression: (sin^(3)θ+cos^(3)θ)/(sinθ+cosθ)

(sin^(3)θ+cos^(3)θ)/(sinθ+cosθ)

[Note : ( A^3 + B^3 ) = (A+B)(A^2-AB+B^2)]

Simplify

= (sinθ + cosθ)(sin^2(θ) - sinθcosθ + cos^(2)θ)

= ( sinθ + cosθ)(sin^2(θ) + cos^(2)θ - sinθcosθ)

[ Formula : (sin^2(θ) + cos^(2)θ = 1 ]

= ( sinθ + cosθ)(1 - sinθcosθ)

There fore

(sin^(3)θ+cos^(3)θ) = ( sinθ + cosθ)(1 - sinθcosθ)

But (sin^(3)θ+cos^(3)θ)/(sinθ+cosθ)

 = ( sinθ + cosθ)(1 - sinθcosθ) / (sinθ+cosθ)

= (1 - sinθcosθ)

There fore

(sin^(3)θ+cos^(3)θ)/(sinθ+cosθ) = (1 - sinθcosθ)