Question
Answer
$$9(1+x)^3-1\cdot(1+x)+6=9x^3+27x^2+26x+14$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}9(1+x)^3-1\cdot(1+x)+6& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}9(1+3x+3x^2+x^3)-1\cdot(1+x)+6 \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}9+27x+27x^2+9x^3-(1+x)+6 \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}9+27x+27x^2+9x^3-1-x+6 \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}9x^3+27x^2+26x+14\end{aligned} $$ | |
| ① | Find $ \left(1+x\right)^3 $ using formula $$ (A + B) = A^3 + 3A^2B + 3AB^2 + B^3 $$where $ A = 1 $ and $ B = x $. $$ \left(1+x\right)^3 = 1^3+3 \cdot 1^2 \cdot x + 3 \cdot 1 \cdot x^2+x^3 = 1+3x+3x^2+x^3 $$ |
| ② | Multiply $ \color{blue}{9} $ by $ \left( 1+3x+3x^2+x^3\right) $ $$ \color{blue}{9} \cdot \left( 1+3x+3x^2+x^3\right) = 9+27x+27x^2+9x^3 $$Multiply $ \color{blue}{1} $ by $ \left( 1+x\right) $ $$ \color{blue}{1} \cdot \left( 1+x\right) = 1+x $$ |
| ③ | Remove the parentheses by changing the sign of each term within them. $$ - \left( 1+x \right) = -1-x $$ |
| ④ | Combine like terms: $$ \color{blue}{9} + \color{red}{27x} +27x^2+9x^3 \color{green}{-1} \color{red}{-x} + \color{green}{6} = 9x^3+27x^2+ \color{red}{26x} + \color{green}{14} $$ |
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