Answer
$$k^2(2k^2-1)+(2k+1)^3=2k^4+8k^3+11k^2+6k+1$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}k^2(2k^2-1)+(2k+1)^3& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}k^2(2k^2-1)+8k^3+12k^2+6k+1 \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}2k^4-k^2+8k^3+12k^2+6k+1 \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}2k^4+8k^3+11k^2+6k+1\end{aligned} $$ | |
| ① | Find $ \left(2k+1\right)^3 $ using formula $$ (A + B) = A^3 + 3A^2B + 3AB^2 + B^3 $$where $ A = 2k $ and $ B = 1 $. $$ \left(2k+1\right)^3 = \left( 2k \right)^3+3 \cdot \left( 2k \right)^2 \cdot 1 + 3 \cdot 2k \cdot 1^2+1^3 = 8k^3+12k^2+6k+1 $$ |
| ② | Multiply $ \color{blue}{k^2} $ by $ \left( 2k^2-1\right) $ $$ \color{blue}{k^2} \cdot \left( 2k^2-1\right) = 2k^4-k^2 $$ |
| ③ | Combine like terms: $$ 2k^4 \color{blue}{-k^2} +8k^3+ \color{blue}{12k^2} +6k+1 = 2k^4+8k^3+ \color{blue}{11k^2} +6k+1 $$ |
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