Factor \( 27 c^{3}-8 \) completely

To factor \( 27c^{3} - 8 \) completely, we first recognize that it is a difference of cubes. The expression can be rewritten as \( (3c)^{3} - 2^{3} \). The formula for factoring a difference of cubes is given by: \[ a^{3} - b^{3} = (a - b)(a^{2} + ab + b^{2}) \] In our case, \( a = 3c \) and \( b = 2 \). Now we will apply the formula: 1. \( a - b = 3c - 2 \) 2. \( a^{2} = (3c)^{2} = 9c^{2} \) 3. \( ab = (3c)(2) = 6c \) 4. \( b^{2} = (2)^{2} = 4 \) Putting it all together: \[ 27c^{3} - 8 = (3c - 2)(9c^{2} + 6c + 4) \] The final factored form of \( 27c^{3} - 8 \) is: \[ (3c - 2)(9c^{2} + 6c + 4) \]