Solve frac{sqrt{2}+sqrt{3}}{sqrt[3]{2}-sqrt{3}}=a-bsqrt{6}

$$\frac { \sqrt { 2 } + \sqrt { 3 } } { \sqrt[ 3 ] { 2 } - \sqrt { 3 } } = a - b \sqrt { 6 }$$

Solve for a (complex solution)

$$a=-\frac{\left(\sqrt{3}+\sqrt[3]{2}\right)\left(\sqrt[6]{864}b-3\sqrt{2}b+\sqrt{2}+\sqrt{3}\right)}{3-2^{\frac{2}{3}}}$$

Solution Steps

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

$$\frac{\left(\sqrt{2}+\sqrt{3}\right)\left(\sqrt[3]{2}+\sqrt{3}\right)}{\left(\sqrt[3]{2}-\sqrt{3}\right)\left(\sqrt[3]{2}+\sqrt{3}\right)}=a-b\sqrt{6}$$

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

$$\frac{\left(\sqrt{2}+\sqrt{3}\right)\left(\sqrt[3]{2}+\sqrt{3}\right)}{\left(\sqrt[3]{2}\right)^{2}-\left(\sqrt{3}\right)^{2}}=a-b\sqrt{6}$$

The square of \sqrt{3} is 3.

$$\frac{\left(\sqrt{2}+\sqrt{3}\right)\left(\sqrt[3]{2}+\sqrt{3}\right)}{\left(\sqrt[3]{2}\right)^{2}-3}=a-b\sqrt{6}$$

Use the distributive property to multiply \sqrt{2}+\sqrt{3} by \sqrt[3]{2}+\sqrt{3}.

$$\frac{\sqrt{2}\sqrt[3]{2}+\sqrt{2}\sqrt{3}+\sqrt{3}\sqrt[3]{2}+\left(\sqrt{3}\right)^{2}}{\left(\sqrt[3]{2}\right)^{2}-3}=a-b\sqrt{6}$$

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

$$\frac{\sqrt{2}\sqrt[3]{2}+\sqrt{6}+\sqrt{3}\sqrt[3]{2}+\left(\sqrt{3}\right)^{2}}{\left(\sqrt[3]{2}\right)^{2}-3}=a-b\sqrt{6}$$

The square of \sqrt{3} is 3.

$$\frac{\sqrt{2}\sqrt[3]{2}+\sqrt{6}+\sqrt{3}\sqrt[3]{2}+3}{\left(\sqrt[3]{2}\right)^{2}-3}=a-b\sqrt{6}$$

Swap sides so that all variable terms are on the left hand side.

$$a-b\sqrt{6}=\frac{\sqrt{2}\sqrt[3]{2}+\sqrt{6}+\sqrt{3}\sqrt[3]{2}+3}{\left(\sqrt[3]{2}\right)^{2}-3}$$

Add b\sqrt{6} to both sides.

$$a=\frac{\sqrt{2}\sqrt[3]{2}+\sqrt{6}+\sqrt{3}\sqrt[3]{2}+3}{\left(\sqrt[3]{2}\right)^{2}-3}+b\sqrt{6}$$

Solve for b (complex solution)

$$b=\frac{\sqrt{6}\left(2\sqrt[3]{2}+3\sqrt[3]{4}+9\right)\left(-2^{\frac{2}{3}}a+3a+\sqrt{3}\sqrt[3]{2}+\sqrt{6}+2^{\frac{5}{6}}+3\right)}{138}$$

Steps for Solving Linear Equation

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

$$\frac{\left(\sqrt{2}+\sqrt{3}\right)\left(\sqrt[3]{2}+\sqrt{3}\right)}{\left(\sqrt[3]{2}-\sqrt{3}\right)\left(\sqrt[3]{2}+\sqrt{3}\right)}=a-b\sqrt{6}$$

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

$$\frac{\left(\sqrt{2}+\sqrt{3}\right)\left(\sqrt[3]{2}+\sqrt{3}\right)}{\left(\sqrt[3]{2}\right)^{2}-\left(\sqrt{3}\right)^{2}}=a-b\sqrt{6}$$

The square of \sqrt{3} is 3.

$$\frac{\left(\sqrt{2}+\sqrt{3}\right)\left(\sqrt[3]{2}+\sqrt{3}\right)}{\left(\sqrt[3]{2}\right)^{2}-3}=a-b\sqrt{6}$$

Use the distributive property to multiply \sqrt{2}+\sqrt{3} by \sqrt[3]{2}+\sqrt{3}.

$$\frac{\sqrt{2}\sqrt[3]{2}+\sqrt{2}\sqrt{3}+\sqrt{3}\sqrt[3]{2}+\left(\sqrt{3}\right)^{2}}{\left(\sqrt[3]{2}\right)^{2}-3}=a-b\sqrt{6}$$

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

$$\frac{\sqrt{2}\sqrt[3]{2}+\sqrt{6}+\sqrt{3}\sqrt[3]{2}+\left(\sqrt{3}\right)^{2}}{\left(\sqrt[3]{2}\right)^{2}-3}=a-b\sqrt{6}$$

The square of \sqrt{3} is 3.

$$\frac{\sqrt{2}\sqrt[3]{2}+\sqrt{6}+\sqrt{3}\sqrt[3]{2}+3}{\left(\sqrt[3]{2}\right)^{2}-3}=a-b\sqrt{6}$$

Swap sides so that all variable terms are on the left hand side.

$$a-b\sqrt{6}=\frac{\sqrt{2}\sqrt[3]{2}+\sqrt{6}+\sqrt{3}\sqrt[3]{2}+3}{\left(\sqrt[3]{2}\right)^{2}-3}$$

Subtract a from both sides.

$$-b\sqrt{6}=\frac{\sqrt{2}\sqrt[3]{2}+\sqrt{6}+\sqrt{3}\sqrt[3]{2}+3}{\left(\sqrt[3]{2}\right)^{2}-3}-a$$

The equation is in standard form.

$$\left(-\sqrt{6}\right)b=-a+\frac{\sqrt{2}\sqrt[3]{2}+\sqrt{3}\sqrt[3]{2}+\sqrt{6}+3}{\left(\sqrt[3]{2}\right)^{2}-3}$$

Divide both sides by -\sqrt{6}.

$$\frac{\left(-\sqrt{6}\right)b}{-\sqrt{6}}=\frac{-2^{\frac{2}{3}}a+3a+\sqrt{3}\sqrt[3]{2}+\sqrt{6}+2^{\frac{5}{6}}+3}{\left(2^{\frac{2}{3}}-3\right)\left(-\sqrt{6}\right)}$$

Dividing by -\sqrt{6} undoes the multiplication by -\sqrt{6}.

$$b=\frac{-2^{\frac{2}{3}}a+3a+\sqrt{3}\sqrt[3]{2}+\sqrt{6}+2^{\frac{5}{6}}+3}{\left(2^{\frac{2}{3}}-3\right)\left(-\sqrt{6}\right)}$$

Divide \frac{2^{\frac{5}{6}}+\sqrt{6}+\sqrt{3}\sqrt[3]{2}+3-a\times 2^{\frac{2}{3}}+3a}{2^{\frac{2}{3}}-3} by -\sqrt{6}.

$$b=-\frac{\sqrt{6}\left(-2^{\frac{2}{3}}a+3a+\sqrt{3}\sqrt[3]{2}+\sqrt{6}+2^{\frac{5}{6}}+3\right)}{6\left(2^{\frac{2}{3}}-3\right)}$$

Solve for a

$$a=-\frac{\left(\sqrt{3}+\sqrt[3]{2}\right)\left(2\sqrt[3]{2}+3\times 2^{\frac{2}{3}}+9\right)\left(\sqrt[6]{864}b-3\sqrt{2}b+\sqrt{2}+\sqrt{3}\right)}{23}$$

Solution Steps

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

$$\frac{\left(\sqrt{2}+\sqrt{3}\right)\left(\sqrt[3]{2}+\sqrt{3}\right)}{\left(\sqrt[3]{2}-\sqrt{3}\right)\left(\sqrt[3]{2}+\sqrt{3}\right)}=a-b\sqrt{6}$$

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

$$\frac{\left(\sqrt{2}+\sqrt{3}\right)\left(\sqrt[3]{2}+\sqrt{3}\right)}{\left(\sqrt[3]{2}\right)^{2}-\left(\sqrt{3}\right)^{2}}=a-b\sqrt{6}$$

The square of \sqrt{3} is 3.

$$\frac{\left(\sqrt{2}+\sqrt{3}\right)\left(\sqrt[3]{2}+\sqrt{3}\right)}{\left(\sqrt[3]{2}\right)^{2}-3}=a-b\sqrt{6}$$

Use the distributive property to multiply \sqrt{2}+\sqrt{3} by \sqrt[3]{2}+\sqrt{3}.

$$\frac{\sqrt{2}\sqrt[3]{2}+\sqrt{2}\sqrt{3}+\sqrt{3}\sqrt[3]{2}+\left(\sqrt{3}\right)^{2}}{\left(\sqrt[3]{2}\right)^{2}-3}=a-b\sqrt{6}$$

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

$$\frac{\sqrt{2}\sqrt[3]{2}+\sqrt{6}+\sqrt{3}\sqrt[3]{2}+\left(\sqrt{3}\right)^{2}}{\left(\sqrt[3]{2}\right)^{2}-3}=a-b\sqrt{6}$$

The square of \sqrt{3} is 3.

$$\frac{\sqrt{2}\sqrt[3]{2}+\sqrt{6}+\sqrt{3}\sqrt[3]{2}+3}{\left(\sqrt[3]{2}\right)^{2}-3}=a-b\sqrt{6}$$

Swap sides so that all variable terms are on the left hand side.

$$a-b\sqrt{6}=\frac{\sqrt{2}\sqrt[3]{2}+\sqrt{6}+\sqrt{3}\sqrt[3]{2}+3}{\left(\sqrt[3]{2}\right)^{2}-3}$$

Add b\sqrt{6} to both sides.

$$a=\frac{\sqrt{2}\sqrt[3]{2}+\sqrt{6}+\sqrt{3}\sqrt[3]{2}+3}{\left(\sqrt[3]{2}\right)^{2}-3}+b\sqrt{6}$$

Solve for b

$$b=\frac{\sqrt{6}\left(\sqrt{3}+\sqrt[3]{2}\right)\left(2\sqrt[3]{2}+3\times 2^{\frac{2}{3}}+9\right)\left(\sqrt{3}a-\sqrt[3]{2}a+\sqrt{2}+\sqrt{3}\right)}{138}$$

Steps for Solving Linear Equation

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

$$\frac{\left(\sqrt{2}+\sqrt{3}\right)\left(\sqrt[3]{2}+\sqrt{3}\right)}{\left(\sqrt[3]{2}-\sqrt{3}\right)\left(\sqrt[3]{2}+\sqrt{3}\right)}=a-b\sqrt{6}$$

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

$$\frac{\left(\sqrt{2}+\sqrt{3}\right)\left(\sqrt[3]{2}+\sqrt{3}\right)}{\left(\sqrt[3]{2}\right)^{2}-\left(\sqrt{3}\right)^{2}}=a-b\sqrt{6}$$

The square of \sqrt{3} is 3.

$$\frac{\left(\sqrt{2}+\sqrt{3}\right)\left(\sqrt[3]{2}+\sqrt{3}\right)}{\left(\sqrt[3]{2}\right)^{2}-3}=a-b\sqrt{6}$$

Use the distributive property to multiply \sqrt{2}+\sqrt{3} by \sqrt[3]{2}+\sqrt{3}.

$$\frac{\sqrt{2}\sqrt[3]{2}+\sqrt{2}\sqrt{3}+\sqrt{3}\sqrt[3]{2}+\left(\sqrt{3}\right)^{2}}{\left(\sqrt[3]{2}\right)^{2}-3}=a-b\sqrt{6}$$

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

$$\frac{\sqrt{2}\sqrt[3]{2}+\sqrt{6}+\sqrt{3}\sqrt[3]{2}+\left(\sqrt{3}\right)^{2}}{\left(\sqrt[3]{2}\right)^{2}-3}=a-b\sqrt{6}$$

The square of \sqrt{3} is 3.

$$\frac{\sqrt{2}\sqrt[3]{2}+\sqrt{6}+\sqrt{3}\sqrt[3]{2}+3}{\left(\sqrt[3]{2}\right)^{2}-3}=a-b\sqrt{6}$$

Swap sides so that all variable terms are on the left hand side.

$$a-b\sqrt{6}=\frac{\sqrt{2}\sqrt[3]{2}+\sqrt{6}+\sqrt{3}\sqrt[3]{2}+3}{\left(\sqrt[3]{2}\right)^{2}-3}$$

Subtract a from both sides.

$$-b\sqrt{6}=\frac{\sqrt{2}\sqrt[3]{2}+\sqrt{6}+\sqrt{3}\sqrt[3]{2}+3}{\left(\sqrt[3]{2}\right)^{2}-3}-a$$

The equation is in standard form.

$$\left(-\sqrt{6}\right)b=-a+\frac{\sqrt{2}\sqrt[3]{2}+\sqrt{3}\sqrt[3]{2}+\sqrt{6}+3}{\left(\sqrt[3]{2}\right)^{2}-3}$$

Divide both sides by -\sqrt{6}.

$$\frac{\left(-\sqrt{6}\right)b}{-\sqrt{6}}=\frac{-2^{\frac{2}{3}}a+3a+\sqrt{3}\sqrt[3]{2}+\sqrt{6}+2^{\frac{5}{6}}+3}{\left(2^{\frac{2}{3}}-3\right)\left(-\sqrt{6}\right)}$$

Dividing by -\sqrt{6} undoes the multiplication by -\sqrt{6}.

$$b=\frac{-2^{\frac{2}{3}}a+3a+\sqrt{3}\sqrt[3]{2}+\sqrt{6}+2^{\frac{5}{6}}+3}{\left(2^{\frac{2}{3}}-3\right)\left(-\sqrt{6}\right)}$$

Divide \frac{2^{\frac{5}{6}}+\sqrt{6}+\sqrt{3}\sqrt[3]{2}+3-a\times 2^{\frac{2}{3}}+3a}{2^{\frac{2}{3}}-3} by -\sqrt{6}.

$$b=-\frac{\sqrt{6}\left(-2^{\frac{2}{3}}a+3a+\sqrt{3}\sqrt[3]{2}+\sqrt{6}+2^{\frac{5}{6}}+3\right)}{6\left(2^{\frac{2}{3}}-3\right)}$$