Solve (2sqrt{7}-5)^2*(2sqrt{7}+5)^2

$$( 2 \sqrt { 7 } - 5 ) ^ { 2 } \cdot ( 2 \sqrt { 7 } + 5 ) ^ { 2 }$$

Evaluate

$$9$$

Solution Steps

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(2\sqrt{7}-5\right)^{2}.

$$\left(4\left(\sqrt{7}\right)^{2}-20\sqrt{7}+25\right)\left(2\sqrt{7}+5\right)^{2}$$

The square of \sqrt{7} is 7.

$$\left(4\times 7-20\sqrt{7}+25\right)\left(2\sqrt{7}+5\right)^{2}$$

Multiply 4 and 7 to get 28.

$$\left(28-20\sqrt{7}+25\right)\left(2\sqrt{7}+5\right)^{2}$$

Add 28 and 25 to get 53.

$$\left(53-20\sqrt{7}\right)\left(2\sqrt{7}+5\right)^{2}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(2\sqrt{7}+5\right)^{2}.

$$\left(53-20\sqrt{7}\right)\left(4\left(\sqrt{7}\right)^{2}+20\sqrt{7}+25\right)$$

The square of \sqrt{7} is 7.

$$\left(53-20\sqrt{7}\right)\left(4\times 7+20\sqrt{7}+25\right)$$

Multiply 4 and 7 to get 28.

$$\left(53-20\sqrt{7}\right)\left(28+20\sqrt{7}+25\right)$$

Add 28 and 25 to get 53.

$$\left(53-20\sqrt{7}\right)\left(53+20\sqrt{7}\right)$$

Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}. Square 53.

$$2809-\left(20\sqrt{7}\right)^{2}$$

Expand \left(20\sqrt{7}\right)^{2}.

$$2809-20^{2}\left(\sqrt{7}\right)^{2}$$

Calculate 20 to the power of 2 and get 400.

$$2809-400\left(\sqrt{7}\right)^{2}$$

The square of \sqrt{7} is 7.

$$2809-400\times 7$$

Multiply 400 and 7 to get 2800.

$$2809-2800$$

Subtract 2800 from 2809 to get 9.

$$9$$

Factor

$$3^{2}$$