Solve (sqrt{3}-1)*(sqrt{3}+1)-(-1/3)^-2+(pi-2)^0+sqrt{8}

$$( \sqrt { 3 } - 1 ) \times ( \sqrt { 3 } + 1 ) - ( - \frac { 1 } { 3 } ) ^ { - 2 } + ( \pi - 2 ) ^ { 0 } + \sqrt { 8 }$$

Evaluate

$$2\sqrt{2}-6\approx -3.171572875$$

Solution Steps

Consider \left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}. Square 1.

$$\left(\sqrt{3}\right)^{2}-1-\left(-\frac{1}{3}\right)^{-2}+\left(\pi -2\right)^{0}+\sqrt{8}$$

The square of \sqrt{3} is 3.

$$3-1-\left(-\frac{1}{3}\right)^{-2}+\left(\pi -2\right)^{0}+\sqrt{8}$$

Subtract 1 from 3 to get 2.

$$2-\left(-\frac{1}{3}\right)^{-2}+\left(\pi -2\right)^{0}+\sqrt{8}$$

Calculate -\frac{1}{3} to the power of -2 and get 9.

$$2-9+\left(\pi -2\right)^{0}+\sqrt{8}$$

Subtract 9 from 2 to get -7.

$$-7+\left(\pi -2\right)^{0}+\sqrt{8}$$

Calculate \pi -2 to the power of 0 and get 1.

$$-7+1+\sqrt{8}$$

Add -7 and 1 to get -6.

$$-6+\sqrt{8}$$

Factor 8=2^{2}\times 2. Rewrite the square root of the product \sqrt{2^{2}\times 2} as the product of square roots \sqrt{2^{2}}\sqrt{2}. Take the square root of 2^{2}.

$$-6+2\sqrt{2}$$