Question
Answer
$$-6ae^{sqrt\cdot5it}-6be^{sqrt-5it}+3ce^{sqrt\cdot2it}+3de^{sqrt-2it}=-6ae^{5iqrst^2}-6be^{sqrt-5it}+3ce^{2iqrst^2}+3de^{sqrt-2it}$$
Explanation
| $$ \begin{aligned}-6ae^{sqrt\cdot5it}-6be^{sqrt-5it}+3ce^{sqrt\cdot2it}+3de^{sqrt-2it}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}-6ae^{5iqrst^2}-6be^{sqrt-5it}+3ce^{2iqrst^2}+3de^{sqrt-2it}\end{aligned} $$ | |
| ① | $$ s q r t \cdot 2 i t = 2 i q r s t^{1 + 1} = 2 i q r s t^2 $$ |
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