Solve (2^3-1)(2^3+1)(2^6+1)(2^12+1)=

$$( 2 ^ { 3 } - 1 ) ( 2 ^ { 3 } + 1 ) ( 2 ^ { 6 } + 1 ) ( 2 ^ { 12 } + 1 ) =$$

Evaluate

$$16777215$$

Solution Steps

Calculate 2 to the power of 3 and get 8.

$$\left(8-1\right)\left(2^{3}+1\right)\left(2^{6}+1\right)\left(2^{12}+1\right)$$

Subtract 1 from 8 to get 7.

$$7\left(2^{3}+1\right)\left(2^{6}+1\right)\left(2^{12}+1\right)$$

Calculate 2 to the power of 3 and get 8.

$$7\left(8+1\right)\left(2^{6}+1\right)\left(2^{12}+1\right)$$

Add 8 and 1 to get 9.

$$7\times 9\left(2^{6}+1\right)\left(2^{12}+1\right)$$

Multiply 7 and 9 to get 63.

$$63\left(2^{6}+1\right)\left(2^{12}+1\right)$$

Calculate 2 to the power of 6 and get 64.

$$63\left(64+1\right)\left(2^{12}+1\right)$$

Add 64 and 1 to get 65.

$$63\times 65\left(2^{12}+1\right)$$

Multiply 63 and 65 to get 4095.

$$4095\left(2^{12}+1\right)$$

Calculate 2 to the power of 12 and get 4096.

$$4095\left(4096+1\right)$$

Add 4096 and 1 to get 4097.

$$4095\times 4097$$

Multiply 4095 and 4097 to get 16777215.

$$16777215$$

Factor

$$3^{2}\times 5\times 7\times 13\times 17\times 241$$