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\left(4\left(\sqrt{7}\right)^{2}-20\sqrt{7}+25\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(4\times 7-20\sqrt{7}+25\right)\left(2\sqrt{7}+5\right)^{2}
The square of \sqrt{7} is 7.
\left(28-20\sqrt{7}+25\right)\left(2\sqrt{7}+5\right)^{2}
Multiply 4 and 7 to get 28.
\left(53-20\sqrt{7}\right)\left(2\sqrt{7}+5\right)^{2}
Add 28 and 25 to get 53.
\left(53-20\sqrt{7}\right)\left(4\left(\sqrt{7}\right)^{2}+20\sqrt{7}+25\right)
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\times 7+20\sqrt{7}+25\right)
The square of \sqrt{7} is 7.
\left(53-20\sqrt{7}\right)\left(28+20\sqrt{7}+25\right)
Multiply 4 and 7 to get 28.
\left(53-20\sqrt{7}\right)\left(53+20\sqrt{7}\right)
Add 28 and 25 to get 53.
2809-\left(20\sqrt{7}\right)^{2}
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.
2809-20^{2}\left(\sqrt{7}\right)^{2}
Expand \left(20\sqrt{7}\right)^{2}.
2809-400\left(\sqrt{7}\right)^{2}
Calculate 20 to the power of 2 and get 400.
2809-400\times 7
The square of \sqrt{7} is 7.
2809-2800
Multiply 400 and 7 to get 2800.
9
Subtract 2800 from 2809 to get 9.