Solve 1/sqrt{10-3}-1/sqrt{10+3}

$$\frac { 1 } { \sqrt { 10 } - 3 } - \frac { 1 } { \sqrt { 10 } + 3 }$$

Evaluate

$$6$$

Solution Steps

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

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

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

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

Square \sqrt{10}. Square 3.

$$\frac{\sqrt{10}+3}{10-9}-\frac{1}{\sqrt{10}+3}$$

Subtract 9 from 10 to get 1.

$$\frac{\sqrt{10}+3}{1}-\frac{1}{\sqrt{10}+3}$$

Anything divided by one gives itself.

$$\sqrt{10}+3-\frac{1}{\sqrt{10}+3}$$

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

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

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

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

Square \sqrt{10}. Square 3.

$$\sqrt{10}+3-\frac{\sqrt{10}-3}{10-9}$$

Subtract 9 from 10 to get 1.

$$\sqrt{10}+3-\frac{\sqrt{10}-3}{1}$$

Anything divided by one gives itself.

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

To find the opposite of \sqrt{10}-3, find the opposite of each term.

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

The opposite of -3 is 3.

$$\sqrt{10}+3-\sqrt{10}+3$$

Combine \sqrt{10} and -\sqrt{10} to get 0.

$$3+3$$

Add 3 and 3 to get 6.

$$6$$

Factor

$$2\times 3$$