Solve (sqrt{3}+1)(sqrt{3}-1)+(sqrt{3}-1)^2

$$( \sqrt { 3 } + 1 ) ( \sqrt { 3 } - 1 ) + ( \sqrt { 3 } - 1 ) ^ { 2 }$$

Evaluate

$$6-2\sqrt{3}\approx 2.535898385$$

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.

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

The square of \sqrt{3} is 3.

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

Subtract 1 from 3 to get 2.

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

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(\sqrt{3}-1\right)^{2}.

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

The square of \sqrt{3} is 3.

$$2+3-2\sqrt{3}+1$$

Add 3 and 1 to get 4.

$$2+4-2\sqrt{3}$$

Add 2 and 4 to get 6.

$$6-2\sqrt{3}$$

Factor

$$2 \sqrt{3} {(\sqrt{3} - 1)} = 2.535898385$$