Answer
$$(6-zi)\cdot(3-i)-(6-zi)\cdot(3-2i)=-i^2z+6i$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}(6-zi)\cdot(3-i)-(6-zi)\cdot(3-2i)& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}18-6i-3iz+i^2z-(18-12i-3iz+2i^2z) \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}18-6i-3iz+i^2z-18+12i+3iz-2i^2z \xlongequal{ } \\[1 em] & \xlongequal{ } \cancel{18}-6i -\cancel{3iz}+i^2z -\cancel{18}+12i+ \cancel{3iz}-2i^2z \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}-i^2z+6i\end{aligned} $$ | |
| ① | Multiply each term of $ \left( \color{blue}{6-iz}\right) $ by each term in $ \left( 3-i\right) $. $$ \left( \color{blue}{6-iz}\right) \cdot \left( 3-i\right) = 18-6i-3iz+i^2z $$Multiply each term of $ \left( \color{blue}{6-iz}\right) $ by each term in $ \left( 3-2i\right) $. $$ \left( \color{blue}{6-iz}\right) \cdot \left( 3-2i\right) = 18-12i-3iz+2i^2z $$ |
| ② | Remove the parentheses by changing the sign of each term within them. $$ - \left( 18-12i-3iz+2i^2z \right) = -18+12i+3iz-2i^2z $$ |
| ③ | Combine like terms: $$ \, \color{blue}{ \cancel{18}} \, \color{green}{-6i} \, \color{orange}{ -\cancel{3iz}} \,+ \color{red}{i^2z} \, \color{blue}{ -\cancel{18}} \,+ \color{green}{12i} + \, \color{orange}{ \cancel{3iz}} \, \color{red}{-2i^2z} = \color{red}{-i^2z} + \color{green}{6i} $$ |
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