Answer
$$\frac{(a+jb)(c+jd)(e+jf)}{g+jh}=\frac{bdfj^3+adfj^2+bcfj^2+bdej^2+acfj+adej+bcej+ace}{g+jh}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{(a+jb)(c+jd)(e+jf)}{g+jh}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{(1ac+adj+bcj+bdj^2)(e+jf)}{g+jh} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{bdfj^3+adfj^2+bcfj^2+bdej^2+acfj+adej+bcej+ace}{g+jh}\end{aligned} $$ | |
| ① | Multiply each term of $ \left( \color{blue}{a+bj}\right) $ by each term in $ \left( c+dj\right) $. $$ \left( \color{blue}{a+bj}\right) \cdot \left( c+dj\right) = ac+adj+bcj+bdj^2 $$ |
| ② | Multiply each term of $ \left( \color{blue}{ac+adj+bcj+bdj^2}\right) $ by each term in $ \left( e+fj\right) $. $$ \left( \color{blue}{ac+adj+bcj+bdj^2}\right) \cdot \left( e+fj\right) = ace+acfj+adej+adfj^2+bcej+bcfj^2+bdej^2+bdfj^3 $$ |
| ③ | Combine like terms: $$ ace+acfj+adej+adfj^2+bcej+bcfj^2+bdej^2+bdfj^3 = bdfj^3+adfj^2+bcfj^2+bdej^2+acfj+adej+bcej+ace $$ |
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