A) \[64-{{x}^{2}}=(64-x)\,(64+x)\]
B) \[27{{x}^{2}}-48=3(3x+4)\,(3x-4)\]
C) \[{{y}^{2}}-81=(y+9)\,(y+9)\]
D) \[36-{{p}^{2}}=(p-6)\,(p+6)\]
Correct Answer: B
Solution :
\[27x{{z}^{2}}-48=3[9{{x}^{2}}-16]\] \[=3\,\,[{{(3x)}^{2}}-{{(4)}^{2}}]\] \[=3\,(3x+4)\,\,(3x-4)\]
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