<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] Hint
$$(a+b+c)^3-c^3=(a+b+c-c)((a+b+c)^2+(a+b+c)c+c^2)$$
$$a^3+b^3=?$$
Clearly $a+b$ is a common factor.
By symmetry, we can claim that the other two factors of $$(a+b+c)^3-a^3-b^3-c^3$$ will be $b+c,c+a$
Alternatively simplify
$$((a+b+c)^2+(a+b+c)c+c^2)-(a^2-ab+b^2)$$
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