<s>[INST] Number of solution of ${\left( {\sin x - 1} \right)^3} + {\left( {\cos x - 1} \right)^3} + {\sin ^3}x = {\left( {2\sin x + \cos x - 2} \right)^3}$ Number of solution of the equation ${\left( {\sin x - 1} \right)^3} + {\left( {\cos x - 1} \right)^3} + {\sin ^3}x = {\left( {2\sin x + \cos x - 2} \right)^3}$ in the interval $[0,2\pi]$ is equal to_____
My approach is as follow
$a = \sin x - 1;b = \cos x - 1;c = \sin x$
${a^3} + {b^3} + {c^3} = {\left( {a + b + c} \right)^3}$
$ \Rightarrow {a^3} + {b^3} + {c^3} = {\left( {a + b + c} \right)^3}$
${\left( {a + b + c} \right)^3} = {a^3} + {b^3} + {c^3} + 6abc + 3ab\left( {a + b} \right) + 3bc\left( {b + c} \right) + 3ac\left( {a + c} \right)$
${a^3} + {b^3} + {c^3} = {\left( {a + b + c} \right)^3} \Rightarrow 6abc + 3ab\left( {a + b} \right) + 3bc\left( {b + c} \right) + 3ac\left( {a + c} \right) = 0$
$ \Rightarrow 2abc + ab\left( {a + b} \right) + bc\left( {b + c} \right) + ac\left( {a + c} \right) = 0$
How do we approach from here
 [/INST] Having $a = \sin x - 1$ and $c = \sin x$ is redundant, it hides potential cancellation and simplification, and leaves you trying to factor a polynomial in three variables instead of just two.  Instead, everywhere you have a $c$, replace it with $a+1$.
\begin{align*}
a^3 + b^3 + (a+1)^3 &= (a+b+a+1)^3  \\
a^3 + b^3 + (a+1)^3 &= (2a+b+1)^3  \\
1 + 3a + 3a^2 + 2a^3 + b^3 &= 8 a^3+12 a^2 b+12 a^2+6 a b^2+12 a b+6 a+b^3+3 b^2+3 b+1  \end{align*}\begin{align*}
-6 a^3-12 a^2 b-9 a^2-6 a b^2-12 a b-3 a-3 b^2-3 b &= 0  \\
-3(2 a^3+4 a^2 b+3 a^2+2 a b^2+4 a b+a+b^2+b) &= 0  \\
-3(2 a^3 + 2 a^2 b + 2 a^2 b + 2 a b^2 + 3 a^2 + 3 a b + a b + b^2 + a+b) &= 0  \\
-3(a+b)(2 a^2 + 2 a b + 3 a + b + 1) &= 0  \\
-3(a+b)(2 a^2 + 2 a b + 2 a + a + b + 1) &= 0  \\
-3(a+b)(a+b+1)(2 a + 1) &= 0
\end{align*}
Note that the factor $a+b+1 = \sin x + \cos x - 1$ leads to three solutions in $[0,2\pi]$.
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