Question
Answer
$$k+1((k+1)^2-1)(3(k-1)+2)-k(k-1)(3k+2)=6k^2+k$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}k+1((k+1)^2-1)(3(k-1)+2)-k(k-1)(3k+2)& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}k+1(1k^2+2k+1-1)(3(k-1)+2)-k(k-1)(3k+2) \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}k+1(1k^2+2k)(3(k-1)+2)-k(k-1)(3k+2) \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}k+(1k^2+2k)(3k-3+2)-(1k^2-k)(3k+2) \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}k+(1k^2+2k)(3k-1)-(1k^2-k)(3k+2) \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle5}{\textcircled {5}} } }}}k+3k^3-k^2+6k^2-2k-(3k^3+2k^2-3k^2-2k) \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle6}{\textcircled {6}} } }}}k+3k^3+5k^2-2k-(3k^3-k^2-2k) \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle7}{\textcircled {7}} } }}}3k^3+5k^2-k-(3k^3-k^2-2k) \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle8}{\textcircled {8}} } }}}3k^3+5k^2-k-3k^3+k^2+2k \xlongequal{ } \\[1 em] & \xlongequal{ } \cancel{3k^3}+5k^2-k -\cancel{3k^3}+k^2+2k \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle9}{\textcircled {9}} } }}}6k^2+k\end{aligned} $$ | |
| ① | Find $ \left(k+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}{ k } $ and $ B = \color{red}{ 1 }$. $$ \begin{aligned}\left(k+1\right)^2 = \color{blue}{k^2} +2 \cdot k \cdot 1 + \color{red}{1^2} = k^2+2k+1\end{aligned} $$ |
| ② | Combine like terms: $$ k^2+2k+ \, \color{blue}{ \cancel{1}} \, \, \color{blue}{ -\cancel{1}} \, = k^2+2k $$ |
| ③ | Multiply $ \color{blue}{1} $ by $ \left( k^2+2k\right) $ $$ \color{blue}{1} \cdot \left( k^2+2k\right) = k^2+2k $$Multiply $ \color{blue}{3} $ by $ \left( k-1\right) $ $$ \color{blue}{3} \cdot \left( k-1\right) = 3k-3 $$Multiply $ \color{blue}{k} $ by $ \left( k-1\right) $ $$ \color{blue}{k} \cdot \left( k-1\right) = k^2-k $$ |
| ④ | Combine like terms: $$ 3k \color{blue}{-3} + \color{blue}{2} = 3k \color{blue}{-1} $$ |
| ⑤ | Multiply each term of $ \left( \color{blue}{k^2+2k}\right) $ by each term in $ \left( 3k-1\right) $. $$ \left( \color{blue}{k^2+2k}\right) \cdot \left( 3k-1\right) = 3k^3-k^2+6k^2-2k $$Multiply each term of $ \left( \color{blue}{k^2-k}\right) $ by each term in $ \left( 3k+2\right) $. $$ \left( \color{blue}{k^2-k}\right) \cdot \left( 3k+2\right) = 3k^3+2k^2-3k^2-2k $$ |
| ⑥ | Combine like terms: $$ 3k^3 \color{blue}{-k^2} + \color{blue}{6k^2} -2k = 3k^3+ \color{blue}{5k^2} -2k $$Combine like terms: $$ 3k^3+ \color{blue}{2k^2} \color{blue}{-3k^2} -2k = 3k^3 \color{blue}{-k^2} -2k $$ |
| ⑦ | Combine like terms: $$ \color{blue}{k} +3k^3+5k^2 \color{blue}{-2k} = 3k^3+5k^2 \color{blue}{-k} $$ |
| ⑧ | Remove the parentheses by changing the sign of each term within them. $$ - \left( 3k^3-k^2-2k \right) = -3k^3+k^2+2k $$ |
| ⑨ | Combine like terms: $$ \, \color{blue}{ \cancel{3k^3}} \,+ \color{green}{5k^2} \color{orange}{-k} \, \color{blue}{ -\cancel{3k^3}} \,+ \color{green}{k^2} + \color{orange}{2k} = \color{green}{6k^2} + \color{orange}{k} $$ |
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