Solve [(25+3):7+(64-4)*10]^2-(3^2-1)^2

$$[ ( 25 + 3 ) : 7 + ( 64 - 4 ) \cdot 10 ] ^ { 2 } - ( 3 ^ { 2 } - 1 ) ^ { 2 }$$

Evaluate

$$364752$$

Solution Steps

Add 25 and 3 to get 28.

$$\left(\frac{28}{7}+\left(64-4\right)\times 10\right)^{2}-\left(3^{2}-1\right)^{2}$$

Divide 28 by 7 to get 4.

$$\left(4+\left(64-4\right)\times 10\right)^{2}-\left(3^{2}-1\right)^{2}$$

Subtract 4 from 64 to get 60.

$$\left(4+60\times 10\right)^{2}-\left(3^{2}-1\right)^{2}$$

Multiply 60 and 10 to get 600.

$$\left(4+600\right)^{2}-\left(3^{2}-1\right)^{2}$$

Add 4 and 600 to get 604.

$$604^{2}-\left(3^{2}-1\right)^{2}$$

Calculate 604 to the power of 2 and get 364816.

$$364816-\left(3^{2}-1\right)^{2}$$

Calculate 3 to the power of 2 and get 9.

$$364816-\left(9-1\right)^{2}$$

Subtract 1 from 9 to get 8.

$$364816-8^{2}$$

Calculate 8 to the power of 2 and get 64.

$$364816-64$$

Subtract 64 from 364816 to get 364752.

$$364752$$

Factor

$$2^{4}\times 3^{2}\times 17\times 149$$