Solve {left(2sqrt{2}right)}^2-1-41+3^-1*6+2^0

$${ \left(2 \sqrt{ 2 } \right) }^{ 2 } -1-41+ { 3 }^{ -1 } \times 6+ { 2 }^{ 0 }$$

Evaluate

$$-31$$

Solution Steps

Expand \left(2\sqrt{2}\right)^{2}.

$$2^{2}\left(\sqrt{2}\right)^{2}-1-41+3^{-1}\times 6+2^{0}$$

Calculate 2 to the power of 2 and get 4.

$$4\left(\sqrt{2}\right)^{2}-1-41+3^{-1}\times 6+2^{0}$$

The square of \sqrt{2} is 2.

$$4\times 2-1-41+3^{-1}\times 6+2^{0}$$

Multiply 4 and 2 to get 8.

$$8-1-41+3^{-1}\times 6+2^{0}$$

Subtract 1 from 8 to get 7.

$$7-41+3^{-1}\times 6+2^{0}$$

Subtract 41 from 7 to get -34.

$$-34+3^{-1}\times 6+2^{0}$$

Calculate 3 to the power of -1 and get \frac{1}{3}.

$$-34+\frac{1}{3}\times 6+2^{0}$$

Multiply \frac{1}{3} and 6 to get 2.

$$-34+2+2^{0}$$

Add -34 and 2 to get -32.

$$-32+2^{0}$$

Calculate 2 to the power of 0 and get 1.

$$-32+1$$

Add -32 and 1 to get -31.

$$-31$$

Factor

$$-31$$