Question

Factor the trigonometric expression and simplify:1-sin^3x

Answer 1

Remember the following rule for factoring a difference of cubes:

$a^3-b^3=(a-b)(a^2+ab+b^2)$

Notice that 1 is equal to 1^3.

Then, to factor the expression:

$1-\sin ^3x$

Use the identity above with a=1 and b=sin(x):

$\begin{gathered} 1-\sin ^3x=(1-\sin x)(1^2+1\cdot\sin x+\sin ^2x) \\ =(1-\sin x)(1+\sin x+\sin ^2x) \end{gathered}$

Therefore, a factored form of the given expression is:

$1-\sin ^3x=(1-\sin x)(1+\sin x+\sin ^2x)$