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 }$.

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

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

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

Multiply each term of $ \left( \color{blue}{t^2+2t+1}\right) $ by each term in $ \left( 4t+1\right) $.

$$ \left( \color{blue}{t^2+2t+1}\right) \cdot \left( 4t+1\right) = 4t^3+t^2+8t^2+2t+4t+1 $$

Combine like terms:

$$ 4t^3+9t^2+6t+1 = 4t^3+9t^2+6t+1 $$

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

$$ - \left( 4t^3+9t^2+6t+1 \right) = -4t^3-9t^2-6t-1 $$

Combine like terms:

$$ \, \color{blue}{ \cancel{4t^3}} \,+ \, \color{green}{ \cancel{9t^2}} \,+ \, \color{blue}{ \cancel{6t}} \,+ \, \color{green}{ \cancel{1}} \, \, \color{blue}{ -\cancel{4t^3}} \, \, \color{green}{ -\cancel{9t^2}} \, \, \color{blue}{ -\cancel{6t}} \, \, \color{green}{ -\cancel{1}} \, = \color{green}{0} $$