Answer
$$(3a-1)^4=81a^4-108a^3+54a^2-12a+1$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}(3a-1)^4& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}81a^4-108a^3+54a^2-12a+1\end{aligned} $$ | |
| ① | $$ (3a-1)^4 = (3a-1)^2 \cdot (3a-1)^2 $$ |
| ② | Find $ \left(3a-1\right)^2 $ using formula. $$ (A - B)^2 = \color{blue}{A^2} - 2 \cdot A \cdot B + \color{red}{B^2} $$where $ A = \color{blue}{ 3a } $ and $ B = \color{red}{ 1 }$. $$ \begin{aligned}\left(3a-1\right)^2 = \color{blue}{\left( 3a \right)^2} -2 \cdot 3a \cdot 1 + \color{red}{1^2} = 9a^2-6a+1\end{aligned} $$ |
| ③ | Multiply each term of $ \left( \color{blue}{9a^2-6a+1}\right) $ by each term in $ \left( 9a^2-6a+1\right) $. $$ \left( \color{blue}{9a^2-6a+1}\right) \cdot \left( 9a^2-6a+1\right) = 81a^4-54a^3+9a^2-54a^3+36a^2-6a+9a^2-6a+1 $$ |
| ④ | Combine like terms: $$ 81a^4 \color{blue}{-54a^3} + \color{red}{9a^2} \color{blue}{-54a^3} + \color{green}{36a^2} \color{orange}{-6a} + \color{green}{9a^2} \color{orange}{-6a} +1 = \\ = 81a^4 \color{blue}{-108a^3} + \color{green}{54a^2} \color{orange}{-12a} +1 $$ |
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