Solve (sqrt{11}-1)^2*(sqrt{11}+1)^2

$$( \sqrt { 11 } - 1 ) ^ { 2 } \cdot ( \sqrt { 11 } + 1 ) ^ { 2 }$$

Evaluate

$$100$$

Solution Steps

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(\sqrt{11}-1\right)^{2}.

$$\left(\left(\sqrt{11}\right)^{2}-2\sqrt{11}+1\right)\left(\sqrt{11}+1\right)^{2}$$

The square of \sqrt{11} is 11.

$$\left(11-2\sqrt{11}+1\right)\left(\sqrt{11}+1\right)^{2}$$

Add 11 and 1 to get 12.

$$\left(12-2\sqrt{11}\right)\left(\sqrt{11}+1\right)^{2}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(\sqrt{11}+1\right)^{2}.

$$\left(12-2\sqrt{11}\right)\left(\left(\sqrt{11}\right)^{2}+2\sqrt{11}+1\right)$$

The square of \sqrt{11} is 11.

$$\left(12-2\sqrt{11}\right)\left(11+2\sqrt{11}+1\right)$$

Add 11 and 1 to get 12.

$$\left(12-2\sqrt{11}\right)\left(12+2\sqrt{11}\right)$$

Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}. Square 12.

$$144-\left(2\sqrt{11}\right)^{2}$$

Expand \left(2\sqrt{11}\right)^{2}.

$$144-2^{2}\left(\sqrt{11}\right)^{2}$$

Calculate 2 to the power of 2 and get 4.

$$144-4\left(\sqrt{11}\right)^{2}$$

The square of \sqrt{11} is 11.

$$144-4\times 11$$

Multiply 4 and 11 to get 44.

$$144-44$$

Subtract 44 from 144 to get 100.

$$100$$

Factor

$$2^{2}\times 5^{2}$$