How do you factor \[{\sin ^3}x - {\cos ^3}x?\]

Hint: The given question describes the operation of using the algebraic formula, arithmetic operations like addition/subtraction/multiplication/division, and trigonometric conditions. Also, it can be solved by comparing the given equation with an algebraic formula. And it involves the substitution of trigonometric equations in solutions.

Complete step by step answer:
The given question is shown below,
\[{\sin ^3}x - {\cos ^3}x\]\[ \to \left( 1 \right)\]
We would factorize the above equation. Let’s see how to factorize the equation\[\left( 1 \right)\],
We know that,
\[{a^3} - {b^3} = \left( {a - b} \right)\left( {{a^2} + {b^2} + ab} \right) \to \left( 2 \right)\]
By comparing the equation\[\left( 1 \right)and\left( 2 \right)\], we get the values \[\left( a \right)and\left( b \right)\]as follows,
\[
  a = \sin x \\
  b = \cos x \\
 \]
So, let’s substitute \[a = \sin x;b = \cos x\] in the equation \[\left( 2 \right)\]we get,
\[\left( {{{\sin }^3}x} \right) - \left( {{{\cos }^3}x} \right) = \left( {\sin x - \cos x} \right)\left( {{{\sin }^2}x + \sin x\cos x + {{\cos }^2}x} \right) \to \left( 3 \right)\]
We know that,
\[{\sin ^2}x + {\cos ^2}x = 1 \to \left( 4 \right)\]
The above equation is used to get a simplified form of solution.
So, we can use “1” instead of\[{\sin ^2}x + {\cos ^2}x\]. So, the equation \[\left( 4 \right)\]is substituted in the equation\[\left( 3 \right)\], we get
\[\left( {{{\sin }^3}x} \right) - \left( {{{\cos }^3}x} \right) = \left( {\sin x - \cos x} \right)\left( {1 + \sin x\cos x} \right) \to \left( 5 \right)\]
Also, we know that
\[\sin 2x = 2\sin x\cos x\] (One of the general trigonometric condition)
\[\dfrac{{\sin 2x}}{2} = \sin x\cos x \to \left( 6 \right)\]
By using the above equation we can get a simplified form of the final solution.
Let’s substitute the equation \[\left( 6 \right)\] in the equation \[\left( 5 \right)\] we get,
\[\left( {{{\sin }^3}x} \right) - \left( {{{\cos }^3}x} \right) = \left( {\sin x - \cos x} \right)(1 + \dfrac{{\sin 2x}}{2}) \to \left( 7 \right)\]
In the next step, we would convert the mixed fraction term into fraction as shown below,
\[\left( {1 + \dfrac{{\sin 2x}}{2}} \right) = \dfrac{{2 + \sin 2x}}{2} \to \left( 8 \right)\]
In the next step, we substituting the equation \[\left( 8 \right)\] in the equation \[\left( 7 \right)\]
\[\left( {{{\sin }^3}x} \right) - \left( {{{\cos }^3}x} \right) = \left( {\sin x - \cos x} \right)(\dfrac{{2 + \sin 2x}}{2})\]
The above equation is the final answer to the given question.

Note:
 In this type of question, we would compare the algebraic formula with the given equation. Also, we would try to substitute trigonometric conditions into the solution as much as possible. We should remember the general trigonometric conditions while solving the problem. At last, the final answer would be the most simplified one.