Solve (10/sqrt{3+1}+1/sqrt{3-2}-6/3-sqrt{3})*(3*sqrt{3}+10)

$$( \frac { 10 } { \sqrt { 3 } + 1 } + \frac { 1 } { \sqrt { 3 } - 2 } - \frac { 6 } { 3 - \sqrt { 3 } } ) \cdot ( 3 \cdot \sqrt { 3 } + 10 )$$

Evaluate

$$-73$$

Solution Steps

Rationalize the denominator of \frac{10}{\sqrt{3}+1} by multiplying numerator and denominator by \sqrt{3}-1.

$$\left(\frac{10\left(\sqrt{3}-1\right)}{\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)}+\frac{1}{\sqrt{3}-2}-\frac{6}{3-\sqrt{3}}\right)\left(3\sqrt{3}+10\right)$$

Consider \left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

$$\left(\frac{10\left(\sqrt{3}-1\right)}{\left(\sqrt{3}\right)^{2}-1^{2}}+\frac{1}{\sqrt{3}-2}-\frac{6}{3-\sqrt{3}}\right)\left(3\sqrt{3}+10\right)$$

Square \sqrt{3}. Square 1.

$$\left(\frac{10\left(\sqrt{3}-1\right)}{3-1}+\frac{1}{\sqrt{3}-2}-\frac{6}{3-\sqrt{3}}\right)\left(3\sqrt{3}+10\right)$$

Subtract 1 from 3 to get 2.

$$\left(\frac{10\left(\sqrt{3}-1\right)}{2}+\frac{1}{\sqrt{3}-2}-\frac{6}{3-\sqrt{3}}\right)\left(3\sqrt{3}+10\right)$$

Divide 10\left(\sqrt{3}-1\right) by 2 to get 5\left(\sqrt{3}-1\right).

$$\left(5\left(\sqrt{3}-1\right)+\frac{1}{\sqrt{3}-2}-\frac{6}{3-\sqrt{3}}\right)\left(3\sqrt{3}+10\right)$$

Rationalize the denominator of \frac{1}{\sqrt{3}-2} by multiplying numerator and denominator by \sqrt{3}+2.

$$\left(5\left(\sqrt{3}-1\right)+\frac{\sqrt{3}+2}{\left(\sqrt{3}-2\right)\left(\sqrt{3}+2\right)}-\frac{6}{3-\sqrt{3}}\right)\left(3\sqrt{3}+10\right)$$

Consider \left(\sqrt{3}-2\right)\left(\sqrt{3}+2\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

$$\left(5\left(\sqrt{3}-1\right)+\frac{\sqrt{3}+2}{\left(\sqrt{3}\right)^{2}-2^{2}}-\frac{6}{3-\sqrt{3}}\right)\left(3\sqrt{3}+10\right)$$

Square \sqrt{3}. Square 2.

$$\left(5\left(\sqrt{3}-1\right)+\frac{\sqrt{3}+2}{3-4}-\frac{6}{3-\sqrt{3}}\right)\left(3\sqrt{3}+10\right)$$

Subtract 4 from 3 to get -1.

$$\left(5\left(\sqrt{3}-1\right)+\frac{\sqrt{3}+2}{-1}-\frac{6}{3-\sqrt{3}}\right)\left(3\sqrt{3}+10\right)$$

Anything divided by -1 gives its opposite. To find the opposite of \sqrt{3}+2, find the opposite of each term.

$$\left(5\left(\sqrt{3}-1\right)-\sqrt{3}-2-\frac{6}{3-\sqrt{3}}\right)\left(3\sqrt{3}+10\right)$$

Rationalize the denominator of \frac{6}{3-\sqrt{3}} by multiplying numerator and denominator by 3+\sqrt{3}.

$$\left(5\left(\sqrt{3}-1\right)-\sqrt{3}-2-\frac{6\left(3+\sqrt{3}\right)}{\left(3-\sqrt{3}\right)\left(3+\sqrt{3}\right)}\right)\left(3\sqrt{3}+10\right)$$

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

$$\left(5\left(\sqrt{3}-1\right)-\sqrt{3}-2-\frac{6\left(3+\sqrt{3}\right)}{3^{2}-\left(\sqrt{3}\right)^{2}}\right)\left(3\sqrt{3}+10\right)$$

Square 3. Square \sqrt{3}.

$$\left(5\left(\sqrt{3}-1\right)-\sqrt{3}-2-\frac{6\left(3+\sqrt{3}\right)}{9-3}\right)\left(3\sqrt{3}+10\right)$$

Subtract 3 from 9 to get 6.

$$\left(5\left(\sqrt{3}-1\right)-\sqrt{3}-2-\frac{6\left(3+\sqrt{3}\right)}{6}\right)\left(3\sqrt{3}+10\right)$$

Cancel out 6 and 6.

$$\left(5\left(\sqrt{3}-1\right)-\sqrt{3}-2-\left(3+\sqrt{3}\right)\right)\left(3\sqrt{3}+10\right)$$

To find the opposite of 3+\sqrt{3}, find the opposite of each term.

$$\left(5\left(\sqrt{3}-1\right)-\sqrt{3}-2-3-\sqrt{3}\right)\left(3\sqrt{3}+10\right)$$

Subtract 3 from -2 to get -5.

$$\left(5\left(\sqrt{3}-1\right)-\sqrt{3}-5-\sqrt{3}\right)\left(3\sqrt{3}+10\right)$$

Combine -\sqrt{3} and -\sqrt{3} to get -2\sqrt{3}.

$$\left(5\left(\sqrt{3}-1\right)-2\sqrt{3}-5\right)\left(3\sqrt{3}+10\right)$$

Use the distributive property to multiply 5 by \sqrt{3}-1.

$$\left(5\sqrt{3}-5-2\sqrt{3}-5\right)\left(3\sqrt{3}+10\right)$$

Combine 5\sqrt{3} and -2\sqrt{3} to get 3\sqrt{3}.

$$\left(3\sqrt{3}-5-5\right)\left(3\sqrt{3}+10\right)$$

Subtract 5 from -5 to get -10.

$$\left(3\sqrt{3}-10\right)\left(3\sqrt{3}+10\right)$$

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

$$\left(3\sqrt{3}\right)^{2}-10^{2}$$

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

$$3^{2}\left(\sqrt{3}\right)^{2}-10^{2}$$

Calculate 3 to the power of 2 and get 9.

$$9\left(\sqrt{3}\right)^{2}-10^{2}$$

The square of \sqrt{3} is 3.

$$9\times 3-10^{2}$$

Multiply 9 and 3 to get 27.

$$27-10^{2}$$

Calculate 10 to the power of 2 and get 100.

$$27-100$$

Subtract 100 from 27 to get -73.

$$-73$$

Factor

$$-73$$