Solve {left(7+4sqrt{3}right)}^2{left(7-4sqrt{3}right)}^2

$${ \left(7+4 \sqrt{ 3 } \right) }^{ 2 } { \left(7-4 \sqrt{ 3 } \right) }^{ 2 }$$

Evaluate

$$1$$

Solution Steps

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(7+4\sqrt{3}\right)^{2}.

$$\left(49+56\sqrt{3}+16\left(\sqrt{3}\right)^{2}\right)\left(7-4\sqrt{3}\right)^{2}$$

The square of \sqrt{3} is 3.

$$\left(49+56\sqrt{3}+16\times 3\right)\left(7-4\sqrt{3}\right)^{2}$$

Multiply 16 and 3 to get 48.

$$\left(49+56\sqrt{3}+48\right)\left(7-4\sqrt{3}\right)^{2}$$

Add 49 and 48 to get 97.

$$\left(97+56\sqrt{3}\right)\left(7-4\sqrt{3}\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(7-4\sqrt{3}\right)^{2}.

$$\left(97+56\sqrt{3}\right)\left(49-56\sqrt{3}+16\left(\sqrt{3}\right)^{2}\right)$$

The square of \sqrt{3} is 3.

$$\left(97+56\sqrt{3}\right)\left(49-56\sqrt{3}+16\times 3\right)$$

Multiply 16 and 3 to get 48.

$$\left(97+56\sqrt{3}\right)\left(49-56\sqrt{3}+48\right)$$

Add 49 and 48 to get 97.

$$\left(97+56\sqrt{3}\right)\left(97-56\sqrt{3}\right)$$

Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}. Square 97.

$$9409-\left(56\sqrt{3}\right)^{2}$$

Expand \left(56\sqrt{3}\right)^{2}.

$$9409-56^{2}\left(\sqrt{3}\right)^{2}$$

Calculate 56 to the power of 2 and get 3136.

$$9409-3136\left(\sqrt{3}\right)^{2}$$

The square of \sqrt{3} is 3.

$$9409-3136\times 3$$

Multiply 3136 and 3 to get 9408.

$$9409-9408$$

Subtract 9408 from 9409 to get 1.

$$1$$

Factor

$$1$$