Giải thích các bước giải:
Đặt `(3+1)(3^2+1)(3^4+1)(3^8+1)(3^16+1)(3^32+1)=A`
`2A=2.(3+1)(3^2+1)(3^4+1)(3^8+1)(3^16+1)(3^32+1)`
`2A=(3-1)(3+1)(3^2+1)(3^4+1)(3^8+1)(3^16+1)(3^32+1)`
`2A=(3^2-1)(3^2+1)(3^4+1)(3^8+1)(3^16+1)(3^32+1)`
`2A=(3^4-1)(3^4+1)(3^8+1)(3^16+1)(3^32+1)`
`2A=(3^8-1)(3^8+1)(3^16+1)(3^32+1)`
`2A=(3^16-1)(3^16+1)(3^32+1)`
`2A=(3^32-1)(3^32+1)`
`2A=3^64-1`
`A=(3^61-1)/2`
$\\$
`\text{Áp dụng hằng đẳng thức :}`
`(a-b).(a+b)=a^2-b^2`
