Question

How do you prove the identity `(sin^3x - cos^3x)/(sinx - cosx) =1 + sinxcosx`?

Answer 1

Start by factoring `sin^3x- cos^3x` using the difference of cubes formula `a^3 - b^3 = (a- b)(a^2 + ab + b^2)`.

So,

`(sin^3x - cos^3x)/(sinx - cosx) = 1 + sinxcosx`

`((sinx - cosx)(sin^2x + sinxcosx + cos^2x))/(sinx - cosx) = 1 + sinxcosx`

The `sinx - cosx` cancel each other out.

`sin^2x + sinxcosx + cos^2x = 1 + sinxcosx`

Use the identity `sin^2theta + cos^2theta = 1`:

`1 + sinxcosx = 1 + sinxcosx`

`LHS = RHS`

Hopefully this helps!

Answer 2

See below.

Explanation:

From the polynomial identity

`(a^3-b^3)/(a-b)=a^2+a b + b^2` we conclude that

`(sin^3x-cos^3x)/(sinx-cosx) = sin^2x+sin x cos x + cos^2x = 1+sinxcosx`