Solve frac{{left(36-18sqrt{3}right)}^2}{{left(81-45sqrt{3}right)}^2}

$$\frac{ { \left(36-18 \sqrt{ 3 } \right) }^{ 2 } }{ { \left(81-45 \sqrt{ 3 } \right) }^{ 2 } }$$

Evaluate

$$\frac{2\left(\sqrt{3}+2\right)}{3}\approx 2.488033872$$

Solution Steps

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(36-18\sqrt{3}\right)^{2}.

$$\frac{1296-1296\sqrt{3}+324\left(\sqrt{3}\right)^{2}}{\left(81-45\sqrt{3}\right)^{2}}$$

The square of \sqrt{3} is 3.

$$\frac{1296-1296\sqrt{3}+324\times 3}{\left(81-45\sqrt{3}\right)^{2}}$$

Multiply 324 and 3 to get 972.

$$\frac{1296-1296\sqrt{3}+972}{\left(81-45\sqrt{3}\right)^{2}}$$

Add 1296 and 972 to get 2268.

$$\frac{2268-1296\sqrt{3}}{\left(81-45\sqrt{3}\right)^{2}}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(81-45\sqrt{3}\right)^{2}.

$$\frac{2268-1296\sqrt{3}}{6561-7290\sqrt{3}+2025\left(\sqrt{3}\right)^{2}}$$

The square of \sqrt{3} is 3.

$$\frac{2268-1296\sqrt{3}}{6561-7290\sqrt{3}+2025\times 3}$$

Multiply 2025 and 3 to get 6075.

$$\frac{2268-1296\sqrt{3}}{6561-7290\sqrt{3}+6075}$$

Add 6561 and 6075 to get 12636.

$$\frac{2268-1296\sqrt{3}}{12636-7290\sqrt{3}}$$

Rationalize the denominator of \frac{2268-1296\sqrt{3}}{12636-7290\sqrt{3}} by multiplying numerator and denominator by 12636+7290\sqrt{3}.

$$\frac{\left(2268-1296\sqrt{3}\right)\left(12636+7290\sqrt{3}\right)}{\left(12636-7290\sqrt{3}\right)\left(12636+7290\sqrt{3}\right)}$$

Consider \left(12636-7290\sqrt{3}\right)\left(12636+7290\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}.

$$\frac{\left(2268-1296\sqrt{3}\right)\left(12636+7290\sqrt{3}\right)}{12636^{2}-\left(-7290\sqrt{3}\right)^{2}}$$

Calculate 12636 to the power of 2 and get 159668496.

$$\frac{\left(2268-1296\sqrt{3}\right)\left(12636+7290\sqrt{3}\right)}{159668496-\left(-7290\sqrt{3}\right)^{2}}$$

Expand \left(-7290\sqrt{3}\right)^{2}.

$$\frac{\left(2268-1296\sqrt{3}\right)\left(12636+7290\sqrt{3}\right)}{159668496-\left(-7290\right)^{2}\left(\sqrt{3}\right)^{2}}$$

Calculate -7290 to the power of 2 and get 53144100.

$$\frac{\left(2268-1296\sqrt{3}\right)\left(12636+7290\sqrt{3}\right)}{159668496-53144100\left(\sqrt{3}\right)^{2}}$$

The square of \sqrt{3} is 3.

$$\frac{\left(2268-1296\sqrt{3}\right)\left(12636+7290\sqrt{3}\right)}{159668496-53144100\times 3}$$

Multiply 53144100 and 3 to get 159432300.

$$\frac{\left(2268-1296\sqrt{3}\right)\left(12636+7290\sqrt{3}\right)}{159668496-159432300}$$

Subtract 159432300 from 159668496 to get 236196.

$$\frac{\left(2268-1296\sqrt{3}\right)\left(12636+7290\sqrt{3}\right)}{236196}$$

Use the distributive property to multiply 2268-1296\sqrt{3} by 12636+7290\sqrt{3} and combine like terms.

$$\frac{28658448+157464\sqrt{3}-9447840\left(\sqrt{3}\right)^{2}}{236196}$$

The square of \sqrt{3} is 3.

$$\frac{28658448+157464\sqrt{3}-9447840\times 3}{236196}$$

Multiply -9447840 and 3 to get -28343520.

$$\frac{28658448+157464\sqrt{3}-28343520}{236196}$$

Subtract 28343520 from 28658448 to get 314928.

$$\frac{314928+157464\sqrt{3}}{236196}$$

Expand

$$\frac{2 \sqrt{3} + 4}{3} = 2.488033871712585$$

Solution Steps

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(36-18\sqrt{3}\right)^{2}.

$$\frac{1296-1296\sqrt{3}+324\left(\sqrt{3}\right)^{2}}{\left(81-45\sqrt{3}\right)^{2}}$$

The square of \sqrt{3} is 3.

$$\frac{1296-1296\sqrt{3}+324\times 3}{\left(81-45\sqrt{3}\right)^{2}}$$

Multiply 324 and 3 to get 972.

$$\frac{1296-1296\sqrt{3}+972}{\left(81-45\sqrt{3}\right)^{2}}$$

Add 1296 and 972 to get 2268.

$$\frac{2268-1296\sqrt{3}}{\left(81-45\sqrt{3}\right)^{2}}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(81-45\sqrt{3}\right)^{2}.

$$\frac{2268-1296\sqrt{3}}{6561-7290\sqrt{3}+2025\left(\sqrt{3}\right)^{2}}$$

The square of \sqrt{3} is 3.

$$\frac{2268-1296\sqrt{3}}{6561-7290\sqrt{3}+2025\times 3}$$

Multiply 2025 and 3 to get 6075.

$$\frac{2268-1296\sqrt{3}}{6561-7290\sqrt{3}+6075}$$

Add 6561 and 6075 to get 12636.

$$\frac{2268-1296\sqrt{3}}{12636-7290\sqrt{3}}$$

Rationalize the denominator of \frac{2268-1296\sqrt{3}}{12636-7290\sqrt{3}} by multiplying numerator and denominator by 12636+7290\sqrt{3}.

$$\frac{\left(2268-1296\sqrt{3}\right)\left(12636+7290\sqrt{3}\right)}{\left(12636-7290\sqrt{3}\right)\left(12636+7290\sqrt{3}\right)}$$

Consider \left(12636-7290\sqrt{3}\right)\left(12636+7290\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}.

$$\frac{\left(2268-1296\sqrt{3}\right)\left(12636+7290\sqrt{3}\right)}{12636^{2}-\left(-7290\sqrt{3}\right)^{2}}$$

Calculate 12636 to the power of 2 and get 159668496.

$$\frac{\left(2268-1296\sqrt{3}\right)\left(12636+7290\sqrt{3}\right)}{159668496-\left(-7290\sqrt{3}\right)^{2}}$$

Expand \left(-7290\sqrt{3}\right)^{2}.

$$\frac{\left(2268-1296\sqrt{3}\right)\left(12636+7290\sqrt{3}\right)}{159668496-\left(-7290\right)^{2}\left(\sqrt{3}\right)^{2}}$$

Calculate -7290 to the power of 2 and get 53144100.

$$\frac{\left(2268-1296\sqrt{3}\right)\left(12636+7290\sqrt{3}\right)}{159668496-53144100\left(\sqrt{3}\right)^{2}}$$

The square of \sqrt{3} is 3.

$$\frac{\left(2268-1296\sqrt{3}\right)\left(12636+7290\sqrt{3}\right)}{159668496-53144100\times 3}$$

Multiply 53144100 and 3 to get 159432300.

$$\frac{\left(2268-1296\sqrt{3}\right)\left(12636+7290\sqrt{3}\right)}{159668496-159432300}$$

Subtract 159432300 from 159668496 to get 236196.

$$\frac{\left(2268-1296\sqrt{3}\right)\left(12636+7290\sqrt{3}\right)}{236196}$$

Use the distributive property to multiply 2268-1296\sqrt{3} by 12636+7290\sqrt{3} and combine like terms.

$$\frac{28658448+157464\sqrt{3}-9447840\left(\sqrt{3}\right)^{2}}{236196}$$

The square of \sqrt{3} is 3.

$$\frac{28658448+157464\sqrt{3}-9447840\times 3}{236196}$$

Multiply -9447840 and 3 to get -28343520.

$$\frac{28658448+157464\sqrt{3}-28343520}{236196}$$

Subtract 28343520 from 28658448 to get 314928.

$$\frac{314928+157464\sqrt{3}}{236196}$$