Find $ \left(9t+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}{ 9t } $ and $ B = \color{red}{ 1 }$.

$$ \begin{aligned}\left(9t+1\right)^2 = \color{blue}{\left( 9t \right)^2} +2 \cdot 9t \cdot 1 + \color{red}{1^2} = 81t^2+18t+1\end{aligned} $$

Find $ \left(3t+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}{ 3t } $ and $ B = \color{red}{ 1 }$.

$$ \begin{aligned}\left(3t+1\right)^2 = \color{blue}{\left( 3t \right)^2} +2 \cdot 3t \cdot 1 + \color{red}{1^2} = 9t^2+6t+1\end{aligned} $$

Multiply each term of $ \left( \color{blue}{9t^2+6t+1}\right) $ by each term in $ \left( 12t+1\right) $.

$$ \left( \color{blue}{9t^2+6t+1}\right) \cdot \left( 12t+1\right) = 108t^3+9t^2+72t^2+6t+12t+1 $$

Combine like terms:

$$ 108t^3+81t^2+18t+1 = 108t^3+81t^2+18t+1 $$

Remove the parentheses by changing the sign of each term within them.

$$ - \left( 108t^3+81t^2+18t+1 \right) = -108t^3-81t^2-18t-1 $$

Combine like terms:

$$ \, \color{blue}{ \cancel{108t^3}} \,+ \, \color{green}{ \cancel{81t^2}} \,+ \, \color{blue}{ \cancel{18t}} \,+ \, \color{green}{ \cancel{1}} \, \, \color{blue}{ -\cancel{108t^3}} \, \, \color{green}{ -\cancel{81t^2}} \, \, \color{blue}{ -\cancel{18t}} \, \, \color{green}{ -\cancel{1}} \, = \color{green}{0} $$