Answer:
[tex] \purple{ \bold{ \therefore \: A_{base} = {x}^{2} + 4x + 16}}[/tex]
Step-by-step explanation:
[tex] \because \: V_{box} = A_{base} \times height \\ \\ \therefore \: A_{base} = \frac{V_{box}}{height} \\ \\ \therefore \: A_{base} = \frac{ {x}^{3} - 64}{(x - 4)} \\ \\ \therefore \: A_{base} = \frac{ {x}^{3} - {4}^{3} }{(x - 4)} \\ \\ \therefore \: A_{base} = \frac{ \cancel{ ({x} - {4})}( {x}^{2} + 4x + {4}^{2} )}{\cancel{ ({x} - {4})}} \\ \\ \purple{ \bold{ \therefore \: A_{base} = {x}^{2} + 4x + 16}}[/tex]
