Solve {left(272sqrt{3}+90right)}^2+{left(146sqrt{3}-180right)}^2+{left(-20sqrt{3}right)}^2

$${ \left(272 \sqrt{ 3 } +90 \right) }^{ 2 } + { \left(146 \sqrt{ 3 } -180 \right) }^{ 2 } + { \left(-20 \sqrt{ 3 } \right) }^{ 2 }$$

Evaluate

$$327600-3600\sqrt{3}\approx 321364.617092752$$

Solution Steps

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(272\sqrt{3}+90\right)^{2}.

$$73984\left(\sqrt{3}\right)^{2}+48960\sqrt{3}+8100+\left(146\sqrt{3}-180\right)^{2}+\left(-20\sqrt{3}\right)^{2}$$

The square of \sqrt{3} is 3.

$$73984\times 3+48960\sqrt{3}+8100+\left(146\sqrt{3}-180\right)^{2}+\left(-20\sqrt{3}\right)^{2}$$

Multiply 73984 and 3 to get 221952.

$$221952+48960\sqrt{3}+8100+\left(146\sqrt{3}-180\right)^{2}+\left(-20\sqrt{3}\right)^{2}$$

Add 221952 and 8100 to get 230052.

$$230052+48960\sqrt{3}+\left(146\sqrt{3}-180\right)^{2}+\left(-20\sqrt{3}\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(146\sqrt{3}-180\right)^{2}.

$$230052+48960\sqrt{3}+21316\left(\sqrt{3}\right)^{2}-52560\sqrt{3}+32400+\left(-20\sqrt{3}\right)^{2}$$

The square of \sqrt{3} is 3.

$$230052+48960\sqrt{3}+21316\times 3-52560\sqrt{3}+32400+\left(-20\sqrt{3}\right)^{2}$$

Multiply 21316 and 3 to get 63948.

$$230052+48960\sqrt{3}[phone]\sqrt{3}+32400+\left(-20\sqrt{3}\right)^{2}$$

Add 63948 and 32400 to get 96348.

$$230052+48960\sqrt{3}[phone]\sqrt{3}+\left(-20\sqrt{3}\right)^{2}$$

Add 230052 and 96348 to get 326400.

$$326400+48960\sqrt{3}-52560\sqrt{3}+\left(-20\sqrt{3}\right)^{2}$$

Combine 48960\sqrt{3} and -52560\sqrt{3} to get -3600\sqrt{3}.

$$326400-3600\sqrt{3}+\left(-20\sqrt{3}\right)^{2}$$

Expand \left(-20\sqrt{3}\right)^{2}.

$$326400-3600\sqrt{3}+\left(-20\right)^{2}\left(\sqrt{3}\right)^{2}$$

Calculate -20 to the power of 2 and get 400.

$$326400-3600\sqrt{3}+400\left(\sqrt{3}\right)^{2}$$

The square of \sqrt{3} is 3.

$$326400-3600\sqrt{3}+400\times 3$$

Multiply 400 and 3 to get 1200.

$$326400-3600\sqrt{3}+1200$$

Add 326400 and 1200 to get 327600.

$$327600-3600\sqrt{3}$$

Expand

$$327600-3600\sqrt{3}$$

Solution Steps

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(272\sqrt{3}+90\right)^{2}.

$$73984\left(\sqrt{3}\right)^{2}+48960\sqrt{3}+8100+\left(146\sqrt{3}-180\right)^{2}+\left(-20\sqrt{3}\right)^{2}$$

The square of \sqrt{3} is 3.

$$73984\times 3+48960\sqrt{3}+8100+\left(146\sqrt{3}-180\right)^{2}+\left(-20\sqrt{3}\right)^{2}$$

Multiply 73984 and 3 to get 221952.

$$221952+48960\sqrt{3}+8100+\left(146\sqrt{3}-180\right)^{2}+\left(-20\sqrt{3}\right)^{2}$$

Add 221952 and 8100 to get 230052.

$$230052+48960\sqrt{3}+\left(146\sqrt{3}-180\right)^{2}+\left(-20\sqrt{3}\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(146\sqrt{3}-180\right)^{2}.

$$230052+48960\sqrt{3}+21316\left(\sqrt{3}\right)^{2}-52560\sqrt{3}+32400+\left(-20\sqrt{3}\right)^{2}$$

The square of \sqrt{3} is 3.

$$230052+48960\sqrt{3}+21316\times 3-52560\sqrt{3}+32400+\left(-20\sqrt{3}\right)^{2}$$

Multiply 21316 and 3 to get 63948.

$$230052+48960\sqrt{3}[phone]\sqrt{3}+32400+\left(-20\sqrt{3}\right)^{2}$$

Add 63948 and 32400 to get 96348.

$$230052+48960\sqrt{3}[phone]\sqrt{3}+\left(-20\sqrt{3}\right)^{2}$$

Add 230052 and 96348 to get 326400.

$$326400+48960\sqrt{3}-52560\sqrt{3}+\left(-20\sqrt{3}\right)^{2}$$

Combine 48960\sqrt{3} and -52560\sqrt{3} to get -3600\sqrt{3}.

$$326400-3600\sqrt{3}+\left(-20\sqrt{3}\right)^{2}$$

Expand \left(-20\sqrt{3}\right)^{2}.

$$326400-3600\sqrt{3}+\left(-20\right)^{2}\left(\sqrt{3}\right)^{2}$$

Calculate -20 to the power of 2 and get 400.

$$326400-3600\sqrt{3}+400\left(\sqrt{3}\right)^{2}$$

The square of \sqrt{3} is 3.

$$326400-3600\sqrt{3}+400\times 3$$

Multiply 400 and 3 to get 1200.

$$326400-3600\sqrt{3}+1200$$

Add 326400 and 1200 to get 327600.

$$327600-3600\sqrt{3}$$