Factor 1-sin^3x

let a = sin x

then we have
1-a^3
which is
(1-a)(1+a+a^2)
so
(1-sin x)(1 + sin x + sin^2 x)

To factor the expression 1-sin^3x, we need to recognize that it is a difference of cubes.

The expression 1-sin^3x can be written as (1-sinx)(1+sin^2x).

Next, we can simplify further using the identity sin^2x = 1-cos^2x:

1+sin^2x = 1+(1-cos^2x) = 1+1-cos^2x = 2-cos^2x.

Therefore, the factored form of the expression 1-sin^3x is (1-sinx)(2-cos^2x).

To factor the expression 1 - sin^3(x), we can use the identity for the cube of a difference:

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

where a = 1 and b = sin(x).

Applying this identity, we have:

1 - sin^3(x) = (1 - sin(x))(1 + sin(x) + sin^2(x)).

Now, let's simplify further. The expression (1 - sin(x)) cannot be simplified any further.

For the expression (1 + sin(x) + sin^2(x)), we can use the Pythagorean identity sin^2(x) + cos^2(x) = 1 to rewrite sin^2(x) as 1 - cos^2(x):

1 + sin(x) + sin^2(x) = 1 + sin(x) + (1 - cos^2(x)).

Combining like terms, we have:

1 + sin(x) + 1 - cos^2(x) = 2 - cos^2(x) + sin(x).

Therefore, the factored form of 1 - sin^3(x) is:

(1 - sin(x))(2 - cos^2(x) + sin(x)).