Answer
$$(n-1)(n^2+n+1)=n^3-1$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}(n-1)(n^2+n+1)& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}n^3+n^2+n-n^2-n-1 \xlongequal{ } \\[1 em] & \xlongequal{ }n^3+ \cancel{n^2}+ \cancel{n} -\cancel{n^2} -\cancel{n}-1 \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}n^3-1\end{aligned} $$ | |
| ① | Multiply each term of $ \left( \color{blue}{n-1}\right) $ by each term in $ \left( n^2+n+1\right) $. $$ \left( \color{blue}{n-1}\right) \cdot \left( n^2+n+1\right) = \\ = n^3+ \cancel{n^2}+ \cancel{n} -\cancel{n^2} -\cancel{n}-1 $$ |
| ② | Combine like terms: $$ n^3+ \, \color{blue}{ \cancel{n^2}} \,+ \, \color{green}{ \cancel{n}} \, \, \color{blue}{ -\cancel{n^2}} \, \, \color{green}{ -\cancel{n}} \,-1 = n^3-1 $$ |
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