Solve {l}{8sqrt{5}+(4-sqrt{5})^2+(5-2sqrt{3})(2sqrt{3}+5)}{(3sqrt{3}-10)^2+60sqrt{3}-(5sqrt{2}-1)(1+5sqrt{2})}

$$\left. \begin{array} { l } { 8 \sqrt { 5 } + ( 4 - \sqrt { 5 } ) ^ { 2 } + ( 5 - 2 \sqrt { 3 } ) ( 2 \sqrt { 3 } + 5 ) } \\ { ( 3 \sqrt { 3 } - 10 ) ^ { 2 } + 60 \sqrt { 3 } - ( 5 \sqrt { 2 } - 1 ) ( 1 + 5 \sqrt { 2 } ) } \end{array} \right.$$

Sort

$$34,\ 78$$

Solution Steps

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(4-\sqrt{5}\right)^{2}.

$$sort(8\sqrt{5}+16-8\sqrt{5}+\left(\sqrt{5}\right)^{2}+\left(5-2\sqrt{3}\right)\left(2\sqrt{3}+5\right),\left(3\sqrt{3}-10\right)^{2}+60\sqrt{3}-\left(5\sqrt{2}-1\right)\left(1+5\sqrt{2}\right))$$

The square of \sqrt{5} is 5.

$$sort(8\sqrt{5}+16-8\sqrt{5}+5+\left(5-2\sqrt{3}\right)\left(2\sqrt{3}+5\right),\left(3\sqrt{3}-10\right)^{2}+60\sqrt{3}-\left(5\sqrt{2}-1\right)\left(1+5\sqrt{2}\right))$$

Add 16 and 5 to get 21.

$$sort(8\sqrt{5}+21-8\sqrt{5}+\left(5-2\sqrt{3}\right)\left(2\sqrt{3}+5\right),\left(3\sqrt{3}-10\right)^{2}+60\sqrt{3}-\left(5\sqrt{2}-1\right)\left(1+5\sqrt{2}\right))$$

Combine 8\sqrt{5} and -8\sqrt{5} to get 0.

$$sort(21+\left(5-2\sqrt{3}\right)\left(2\sqrt{3}+5\right),\left(3\sqrt{3}-10\right)^{2}+60\sqrt{3}-\left(5\sqrt{2}-1\right)\left(1+5\sqrt{2}\right))$$

Consider \left(5-2\sqrt{3}\right)\left(2\sqrt{3}+5\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 5.

$$sort(21+25-\left(2\sqrt{3}\right)^{2},\left(3\sqrt{3}-10\right)^{2}+60\sqrt{3}-\left(5\sqrt{2}-1\right)\left(1+5\sqrt{2}\right))$$

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

$$sort(21+25-2^{2}\left(\sqrt{3}\right)^{2},\left(3\sqrt{3}-10\right)^{2}+60\sqrt{3}-\left(5\sqrt{2}-1\right)\left(1+5\sqrt{2}\right))$$

Calculate 2 to the power of 2 and get 4.

$$sort(21+25-4\left(\sqrt{3}\right)^{2},\left(3\sqrt{3}-10\right)^{2}+60\sqrt{3}-\left(5\sqrt{2}-1\right)\left(1+5\sqrt{2}\right))$$

The square of \sqrt{3} is 3.

$$sort(21+25-4\times 3,\left(3\sqrt{3}-10\right)^{2}+60\sqrt{3}-\left(5\sqrt{2}-1\right)\left(1+5\sqrt{2}\right))$$

Multiply 4 and 3 to get 12.

$$sort(21+25-12,\left(3\sqrt{3}-10\right)^{2}+60\sqrt{3}-\left(5\sqrt{2}-1\right)\left(1+5\sqrt{2}\right))$$

Subtract 12 from 25 to get 13.

$$sort(21+13,\left(3\sqrt{3}-10\right)^{2}+60\sqrt{3}-\left(5\sqrt{2}-1\right)\left(1+5\sqrt{2}\right))$$

Add 21 and 13 to get 34.

$$sort(34,\left(3\sqrt{3}-10\right)^{2}+60\sqrt{3}-\left(5\sqrt{2}-1\right)\left(1+5\sqrt{2}\right))$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(3\sqrt{3}-10\right)^{2}.

$$sort(34,9\left(\sqrt{3}\right)^{2}-60\sqrt{3}+100+60\sqrt{3}-\left(5\sqrt{2}-1\right)\left(1+5\sqrt{2}\right))$$

The square of \sqrt{3} is 3.

$$sort(34,9\times 3-60\sqrt{3}+100+60\sqrt{3}-\left(5\sqrt{2}-1\right)\left(1+5\sqrt{2}\right))$$

Multiply 9 and 3 to get 27.

$$sort(34,27-60\sqrt{3}+100+60\sqrt{3}-\left(5\sqrt{2}-1\right)\left(1+5\sqrt{2}\right))$$

Add 27 and 100 to get 127.

$$sort(34,127-60\sqrt{3}+60\sqrt{3}-\left(5\sqrt{2}-1\right)\left(1+5\sqrt{2}\right))$$

Combine -60\sqrt{3} and 60\sqrt{3} to get 0.

$$sort(34,127-\left(5\sqrt{2}-1\right)\left(1+5\sqrt{2}\right))$$

Use the distributive property to multiply 5\sqrt{2}-1 by 1+5\sqrt{2} and combine like terms.

$$sort(34,127-\left(25\left(\sqrt{2}\right)^{2}-1\right))$$

The square of \sqrt{2} is 2.

$$sort(34,127-\left(25\times 2-1\right))$$

Multiply 25 and 2 to get 50.

$$sort(34,127-\left(50-1\right))$$

Subtract 1 from 50 to get 49.

$$sort(34,127-49)$$

Subtract 49 from 127 to get 78.

$$sort(34,78)$$

The list values are already in order.

$$34,78$$

Evaluate

$$34,\ 78$$