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Identifying the given you'll have:
 the change of volume per time,
      dV/dt = 6cm/min
 the length of a side      
      s = 12cm
Volume of a cube, V = s^{2}
   differentiating it with time,
   \frac{dV}{dt} = 3s²  \frac{ds}{dt}
 substituting the values given:
  6 = 3(12)² \frac{ds}{dt}
 \frac{ds}{dt} = 0.0139 cm/min
0 0 0
Are you able to understand how it's done?
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