please explain 6

There is a mistake in your question.
So, assuming your question as
Prove that both the sides of (1-sinA)(1-sinB)(1-sinC)=(1+sinA)(1+sinB)(1+sinC)= ±  cosA cosB cosC
$$\begin{gathered} \left(1-\mathrm{sinA}\right)\left(1-\mathrm{sinB}\right)\left(1-\mathrm{sinC}\right)=\left(1+\mathrm{sinA}\right)\left(1+\mathrm{sinB}\right)\left(1+\mathrm{sinC}\right) \\ \mathrm{Mutiplying} \mathrm{by} \left(1+\mathrm{sinA}\right)\left(1+\mathrm{sinB}\right)\left(1+\mathrm{sinC}\right) \mathrm{we} \mathrm{get}, \\ \left(1-\mathrm{sinA}\right)\left(1-\mathrm{sinB}\right)\left(1-\mathrm{sinC}\right)\left(1+\mathrm{sinA}\right)\left(1+\mathrm{sinB}\right)\left(1+\mathrm{sinC}\right)=\left(1+\mathrm{sinA}\right)\left(1+\mathrm{sinB}\right)\left(1+\mathrm{sinC}\right)\left(1+\mathrm{sinA}\right)\left(1+\mathrm{sinB}\right)\left(1+\mathrm{sinC}\right) \\ \left(1-{\sin }^2\mathrm{A}\right)\left(1-{\sin }^2\mathrm{B}\right)\left(1-{\sin }^2\mathrm{C}\right)={\left[\left(1+\mathrm{sinA}\right)\left(1+\mathrm{sinB}\right)\left(1+\mathrm{sinC}\right)\right]}^2 \\ {\cos }^2\mathrm{A}\times {\cos }^2\mathrm{B}\times {\cos }^2\mathrm{C}={\left[\left(1+\mathrm{sinA}\right)\left(1+\mathrm{sinB}\right)\left(1+\mathrm{sinC}\right)\right]}^2 \\ \left(1+\mathrm{sinA}\right)\left(1+\mathrm{sinB}\right)\left(1+\mathrm{sinC}\right)=\sqrt{{\cos }^2\mathrm{A}\times {\cos }^2\mathrm{B}\times {\cos }^2\mathrm{C}}=\pm \mathrm{cosAcosBcosC} \\ \mathrm{So} \left(1-\mathrm{sinA}\right)\left(1-\mathrm{sinB}\right)\left(1-\mathrm{sinC}\right)=\left(1+\mathrm{sinA}\right)\left(1+\mathrm{sinB}\right)\left(1+\mathrm{sinC}\right)=\pm \mathrm{cosAcosBcosC} \end{gathered}$$
Note: sin²x+cos²x=1
and (a²-b²)=(a-b)(a+b)