Answer
$$(3+3y)^3=27y^3+81y^2+81y+27$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}(3+3y)^3& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}27+81y+81y^2+27y^3 \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}27y^3+81y^2+81y+27\end{aligned} $$ | |
| ① | Find $ \left(3+3y\right)^3 $ using formula $$ (A + B) = A^3 + 3A^2B + 3AB^2 + B^3 $$where $ A = 3 $ and $ B = 3y $. $$ \left(3+3y\right)^3 = 3^3+3 \cdot 3^2 \cdot 3y + 3 \cdot 3 \cdot \left( 3y \right)^2+\left( 3y \right)^3 = 27+81y+81y^2+27y^3 $$ |
| ② | Combine like terms: $$ 27y^3+81y^2+81y+27 = 27y^3+81y^2+81y+27 $$ |
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