Lời giải:
Áp dụng hằng đẳng thức \((a-1)(a+1)=a^2-1\) ta có:
\(A=3(2^2+1)(2^4+1)(2^8+1)(2^{16}+1)\)
\(=(2^2-1)(2^2+1)(2^4+1)(2^8+1)(2^{16}+1)\)
\(=(2^4-1)(2^4+1)(2^8+1)(2^{16}+1)\)
\(=(2^8-1)(2^8+1)(2^{16}+1)\)
\(=(2^{16}-1)(2^{16}+1)=2^{32}-1\)