Question
Answer
$$(n^4+3n^2+1)^2-(n^4+n^2+1)=n^8+6n^6+10n^4+5n^2$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}(n^4+3n^2+1)^2-(n^4+n^2+1)& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}n^8+6n^6+11n^4+6n^2+1-(n^4+n^2+1) \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}n^8+6n^6+11n^4+6n^2+1-n^4-n^2-1 \xlongequal{ } \\[1 em] & \xlongequal{ }n^8+6n^6+11n^4+6n^2+ \cancel{1}-n^4-n^2 -\cancel{1} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}n^8+6n^6+10n^4+5n^2\end{aligned} $$ | |
| ① | Multiply each term of $ \left( \color{blue}{n^4+3n^2+1}\right) $ by each term in $ \left( n^4+3n^2+1\right) $. $$ \left( \color{blue}{n^4+3n^2+1}\right) \cdot \left( n^4+3n^2+1\right) = n^8+3n^6+n^4+3n^6+9n^4+3n^2+n^4+3n^2+1 $$ |
| ② | Combine like terms: $$ n^8+ \color{blue}{3n^6} + \color{red}{n^4} + \color{blue}{3n^6} + \color{green}{9n^4} + \color{orange}{3n^2} + \color{green}{n^4} + \color{orange}{3n^2} +1 = \\ = n^8+ \color{blue}{6n^6} + \color{green}{11n^4} + \color{orange}{6n^2} +1 $$ |
| ③ | Remove the parentheses by changing the sign of each term within them. $$ - \left( n^4+n^2+1 \right) = -n^4-n^2-1 $$ |
| ④ | Combine like terms: $$ n^8+6n^6+ \color{blue}{11n^4} + \color{red}{6n^2} + \, \color{green}{ \cancel{1}} \, \color{blue}{-n^4} \color{red}{-n^2} \, \color{green}{ -\cancel{1}} \, = n^8+6n^6+ \color{blue}{10n^4} + \color{red}{5n^2} $$ |
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