Solve {l}{x=1-sqrt{2}}{y={(x-1/x)}^3}{text{Solvefor}ztext{where}}{z=y}

$$\left. \begin{array} { l } { x = 1 - \sqrt{2} }\\ { y = {(x - \frac{1}{x})} ^ {3} }\\ { \text{Solve for } z \text{ where} } \\ { z = y } \end{array} \right.$$

Solve for x, y, z

$$z=8$$

Solution Steps

Consider the second equation. Insert the known values of variables into the equation.

$$y=\left(1-\sqrt{2}-\frac{1}{1-\sqrt{2}}\right)^{3}$$

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

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

Consider \left(1-\sqrt{2}\right)\left(1+\sqrt{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}.

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

Square 1. Square \sqrt{2}.

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

Subtract 2 from 1 to get -1.

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

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

$$y=\left(1-\sqrt{2}-\left(-1-\sqrt{2}\right)\right)^{3}$$

To find the opposite of -1-\sqrt{2}, find the opposite of each term.

$$y=\left(1-\sqrt{2}+1+\sqrt{2}\right)^{3}$$

Add 1 and 1 to get 2.

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

Combine -\sqrt{2} and \sqrt{2} to get 0.

$$y=2^{3}$$

Calculate 2 to the power of 3 and get 8.

$$y=8$$

Consider the third equation. Insert the known values of variables into the equation.

$$z=8$$

The system is now solved.

$$x=1-\sqrt{2} y=8 z=8$$