Question
Answer
$$(1-a-4b-3c)^2=a^2+8ab+6ac+16b^2+24bc+9c^2-2a-8b-6c+1$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}(1-a-4b-3c)^2& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}a^2+8ab+6ac+16b^2+24bc+9c^2-2a-8b-6c+1\end{aligned} $$ | |
| ① | Multiply each term of $ \left( \color{blue}{1-a-4b-3c}\right) $ by each term in $ \left( 1-a-4b-3c\right) $. $$ \left( \color{blue}{1-a-4b-3c}\right) \cdot \left( 1-a-4b-3c\right) = \\ = 1-a-4b-3c-a+a^2+4ab+3ac-4b+4ab+16b^2+12bc-3c+3ac+12bc+9c^2 $$ |
| ② | Combine like terms: $$ 1 \color{blue}{-a} \color{red}{-4b} \color{green}{-3c} \color{blue}{-a} +a^2+ \color{orange}{4ab} + \color{blue}{3ac} \color{red}{-4b} + \color{orange}{4ab} +16b^2+ \color{red}{12bc} \color{green}{-3c} + \color{blue}{3ac} + \color{red}{12bc} +9c^2 = \\ = a^2+ \color{orange}{8ab} + \color{blue}{6ac} +16b^2+ \color{red}{24bc} +9c^2 \color{blue}{-2a} \color{red}{-8b} \color{green}{-6c} +1 $$ |
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