Solve (2/sqrt{2}+3/sqrt{3})(sqrt{3}+sqrt{2}7)(sqrt{2}-sqrt{3})

$$( \frac { 2 } { \sqrt { 2 } } + \frac { 3 } { \sqrt { 3 } } ) ( \sqrt { 3 } + \sqrt { 2 } 7 ) ( \sqrt { 2 } - \sqrt { 3 } )$$

Evaluate

$$-\sqrt{3}-7\sqrt{2}\approx -11.631545744$$

Solution Steps

Rationalize the denominator of \frac{2}{\sqrt{2}} by multiplying numerator and denominator by \sqrt{2}.

$$\left(\frac{2\sqrt{2}}{\left(\sqrt{2}\right)^{2}}+\frac{3}{\sqrt{3}}\right)\left(\sqrt{3}+7\sqrt{2}\right)\left(\sqrt{2}-\sqrt{3}\right)$$

The square of \sqrt{2} is 2.

$$\left(\frac{2\sqrt{2}}{2}+\frac{3}{\sqrt{3}}\right)\left(\sqrt{3}+7\sqrt{2}\right)\left(\sqrt{2}-\sqrt{3}\right)$$

Cancel out 2 and 2.

$$\left(\sqrt{2}+\frac{3}{\sqrt{3}}\right)\left(\sqrt{3}+7\sqrt{2}\right)\left(\sqrt{2}-\sqrt{3}\right)$$

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

$$\left(\sqrt{2}+\frac{3\sqrt{3}}{\left(\sqrt{3}\right)^{2}}\right)\left(\sqrt{3}+7\sqrt{2}\right)\left(\sqrt{2}-\sqrt{3}\right)$$

The square of \sqrt{3} is 3.

$$\left(\sqrt{2}+\frac{3\sqrt{3}}{3}\right)\left(\sqrt{3}+7\sqrt{2}\right)\left(\sqrt{2}-\sqrt{3}\right)$$

Cancel out 3 and 3.

$$\left(\sqrt{2}+\sqrt{3}\right)\left(\sqrt{3}+7\sqrt{2}\right)\left(\sqrt{2}-\sqrt{3}\right)$$

Apply the distributive property by multiplying each term of \sqrt{2}+\sqrt{3} by each term of \sqrt{3}+7\sqrt{2}.

$$\left(\sqrt{2}\sqrt{3}+7\left(\sqrt{2}\right)^{2}+\left(\sqrt{3}\right)^{2}+7\sqrt{3}\sqrt{2}\right)\left(\sqrt{2}-\sqrt{3}\right)$$

To multiply \sqrt{2} and \sqrt{3}, multiply the numbers under the square root.

$$\left(\sqrt{6}+7\left(\sqrt{2}\right)^{2}+\left(\sqrt{3}\right)^{2}+7\sqrt{3}\sqrt{2}\right)\left(\sqrt{2}-\sqrt{3}\right)$$

The square of \sqrt{2} is 2.

$$\left(\sqrt{6}+7\times 2+\left(\sqrt{3}\right)^{2}+7\sqrt{3}\sqrt{2}\right)\left(\sqrt{2}-\sqrt{3}\right)$$

Multiply 7 and 2 to get 14.

$$\left(\sqrt{6}+14+\left(\sqrt{3}\right)^{2}+7\sqrt{3}\sqrt{2}\right)\left(\sqrt{2}-\sqrt{3}\right)$$

The square of \sqrt{3} is 3.

$$\left(\sqrt{6}+14+3+7\sqrt{3}\sqrt{2}\right)\left(\sqrt{2}-\sqrt{3}\right)$$

Add 14 and 3 to get 17.

$$\left(\sqrt{6}+17+7\sqrt{3}\sqrt{2}\right)\left(\sqrt{2}-\sqrt{3}\right)$$

To multiply \sqrt{3} and \sqrt{2}, multiply the numbers under the square root.

$$\left(\sqrt{6}+17+7\sqrt{6}\right)\left(\sqrt{2}-\sqrt{3}\right)$$

Combine \sqrt{6} and 7\sqrt{6} to get 8\sqrt{6}.

$$\left(8\sqrt{6}+17\right)\left(\sqrt{2}-\sqrt{3}\right)$$

Apply the distributive property by multiplying each term of 8\sqrt{6}+17 by each term of \sqrt{2}-\sqrt{3}.

$$8\sqrt{6}\sqrt{2}-8\sqrt{3}\sqrt{6}+17\sqrt{2}-17\sqrt{3}$$

Factor 6=2\times 3. Rewrite the square root of the product \sqrt{2\times 3} as the product of square roots \sqrt{2}\sqrt{3}.

$$8\sqrt{2}\sqrt{3}\sqrt{2}-8\sqrt{3}\sqrt{6}+17\sqrt{2}-17\sqrt{3}$$

Multiply \sqrt{2} and \sqrt{2} to get 2.

$$8\times 2\sqrt{3}-8\sqrt{3}\sqrt{6}+17\sqrt{2}-17\sqrt{3}$$

Multiply 8 and 2 to get 16.

$$16\sqrt{3}-8\sqrt{3}\sqrt{6}+17\sqrt{2}-17\sqrt{3}$$

Factor 6=3\times 2. Rewrite the square root of the product \sqrt{3\times 2} as the product of square roots \sqrt{3}\sqrt{2}.

$$16\sqrt{3}-8\sqrt{3}\sqrt{3}\sqrt{2}+17\sqrt{2}-17\sqrt{3}$$

Multiply \sqrt{3} and \sqrt{3} to get 3.

$$16\sqrt{3}-8\times 3\sqrt{2}+17\sqrt{2}-17\sqrt{3}$$

Multiply -8 and 3 to get -24.

$$16\sqrt{3}-24\sqrt{2}+17\sqrt{2}-17\sqrt{3}$$

Combine -24\sqrt{2} and 17\sqrt{2} to get -7\sqrt{2}.

$$16\sqrt{3}-7\sqrt{2}-17\sqrt{3}$$

Combine 16\sqrt{3} and -17\sqrt{3} to get -\sqrt{3}.

$$-\sqrt{3}-7\sqrt{2}$$