Two variable quantities $A$ and $B$ are found to be related by the equation given below. What is the rate of change $d A / d t$ at the moment when $A=2$ and $d B / d t=1$ ? \[ A^{3}+B^{3}=133 \] \[ \frac{d A}{d t}=\square \text { when } A=2 \text { and } d B / d t=1 \] (Simplify your answer.)

Step 1 ：Given the equation \(A^{3}+B^{3}=133\), we can differentiate both sides with respect to \(t\).

Step 2 ：This gives us \(\frac{d}{dt}(A^{3}+B^{3})=\frac{d}{dt}(133)\).

Step 3 ：Applying the chain rule, we get \(3A^{2}\frac{dA}{dt}+3B^{2}\frac{dB}{dt}=0\).

Step 4 ：We are given that \(A=2\) and \(\frac{dB}{dt}=1\). We can substitute these values into the equation.

Step 5 ：This gives us \(3(2)^{2}\frac{dA}{dt}+3B^{2}(1)=0\).

Step 6 ：We can solve the original equation \(A^{3}+B^{3}=133\) for \(B\) when \(A=2\).

Step 7 ：This gives us \(B=\sqrt[3]{133-2^{3}}\).

Step 8 ：Substitute this value of \(B\) into the equation to get \(3(2)^{2}\frac{dA}{dt}+3(\sqrt[3]{133-2^{3}})^{2}(1)=0\).

Step 9 ：Solving this equation for \(\frac{dA}{dt}\), we get \(\frac{dA}{dt}=-\frac{(\sqrt[3]{133-2^{3}})^{2}}{4}\).

Step 10 ：So, the rate of change \(dA/dt\) at the moment when \(A=2\) and \(dB/dt=1\) is \(\boxed{-\frac{(\sqrt[3]{133-2^{3}})^{2}}{4}}\).