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

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

Solve for b

$$b=\frac{\sqrt{6}}{3}+\frac{2\sqrt[6]{3}}{573}+\frac{4\sqrt[3]{3}}{573}+\frac{8\sqrt{3}}{573}+\frac{16\times 3^{\frac{2}{3}}}{573}+\frac{32\times 3^{\frac{5}{6}}}{573}+\frac{64}{191}\approx 1.387606886$$

Steps for Solving Linear Equation

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

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

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

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

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

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

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

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

Calculate -2 to the power of 2 and get 4.

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

The square of \sqrt{3} is 3.

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

Multiply 4 and 3 to get 12.

$$\frac{2\sqrt{2}\left(\sqrt[3]{3}+2\sqrt{3}\right)}{\left(\sqrt[3]{3}\right)^{2}-12}=2-b\sqrt{6}$$

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

$$\frac{2\sqrt{2}\sqrt[3]{3}+4\sqrt{3}\sqrt{2}}{\left(\sqrt[3]{3}\right)^{2}-12}=2-b\sqrt{6}$$

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

$$\frac{2\sqrt{2}\sqrt[3]{3}+4\sqrt{6}}{\left(\sqrt[3]{3}\right)^{2}-12}=2-b\sqrt{6}$$

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

$$2-b\sqrt{6}=\frac{2\sqrt{2}\sqrt[3]{3}+4\sqrt{6}}{\left(\sqrt[3]{3}\right)^{2}-12}$$

Subtract 2 from both sides.

$$-b\sqrt{6}=\frac{2\sqrt{2}\sqrt[3]{3}+4\sqrt{6}}{\left(\sqrt[3]{3}\right)^{2}-12}-2$$

The equation is in standard form.

$$\left(-\sqrt{6}\right)b=\frac{2\sqrt{2}\sqrt[3]{3}+4\sqrt{6}}{\left(\sqrt[3]{3}\right)^{2}-12}-2$$

Divide both sides by -\sqrt{6}.

$$\frac{\left(-\sqrt{6}\right)b}{-\sqrt{6}}=\frac{2\left(\sqrt{2}\sqrt[3]{3}+2\sqrt{6}+12-3^{\frac{2}{3}}\right)}{\left(3^{\frac{2}{3}}-12\right)\left(-\sqrt{6}\right)}$$

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

$$b=\frac{2\left(\sqrt{2}\sqrt[3]{3}+2\sqrt{6}+12-3^{\frac{2}{3}}\right)}{\left(3^{\frac{2}{3}}-12\right)\left(-\sqrt{6}\right)}$$

Divide \frac{2\left(\sqrt{2}\sqrt[3]{3}+2\sqrt{6}-3^{\frac{2}{3}}+12\right)}{3^{\frac{2}{3}}-12} by -\sqrt{6}.

$$b=-\frac{\sqrt{6}\left(\sqrt{2}\sqrt[3]{3}+2\sqrt{6}+12-3^{\frac{2}{3}}\right)}{3\left(3^{\frac{2}{3}}-12\right)}$$