Solve {c}{x/3+3/x=1}{text{then}}{x^3=?}

$$\left. \begin{array} { c } { \frac { x } { 3 } + \frac { 3 } { x } = 1 } \\ { \text { then } } \\ { x ^ { 3 } = ? } \end{array} \right.$$

Solve for x

$$x=\frac{3+3\sqrt{3}i}{2}\approx 1.5+2.598076211i$$

$$x=\frac{-3\sqrt{3}i+3}{2}\approx 1.5-2.598076211i$$

Steps Using the Quadratic Formula

Variable x cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by 3x, the least common multiple of 3,x.

$$xx+3\times 3=3x$$

Multiply x and x to get x^{2}.

$$x^{2}+3\times 3=3x$$

Multiply 3 and 3 to get 9.

$$x^{2}+9=3x$$

Subtract 3x from both sides.

$$x^{2}+9-3x=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

$$x^{2}-3x+9=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -3 for b, and 9 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-3\right)±\sqrt{\left(-3\right)^{2}-4\times 9}}{2}$$

Square -3.

$$x=\frac{-\left(-3\right)±\sqrt{9-4\times 9}}{2}$$

Multiply -4 times 9.

$$x=\frac{-\left(-3\right)±\sqrt{9-36}}{2}$$

Add 9 to -36.

$$x=\frac{-\left(-3\right)±\sqrt{-27}}{2}$$

Take the square root of -27.

$$x=\frac{-\left(-3\right)±3\sqrt{3}i}{2}$$

The opposite of -3 is 3.

$$x=\frac{3±3\sqrt{3}i}{2}$$

Now solve the equation x=\frac{3±3\sqrt{3}i}{2} when ± is plus. Add 3 to 3i\sqrt{3}.

$$x=\frac{3+3\sqrt{3}i}{2}$$

Now solve the equation x=\frac{3±3\sqrt{3}i}{2} when ± is minus. Subtract 3i\sqrt{3} from 3.

$$x=\frac{-3\sqrt{3}i+3}{2}$$

The equation is now solved.

$$x=\frac{3+3\sqrt{3}i}{2} x=\frac{-3\sqrt{3}i+3}{2}$$

Steps for Completing the Square

Variable x cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by 3x, the least common multiple of 3,x.

$$xx+3\times 3=3x$$

Multiply x and x to get x^{2}.

$$x^{2}+3\times 3=3x$$

Multiply 3 and 3 to get 9.

$$x^{2}+9=3x$$

Subtract 3x from both sides.

$$x^{2}+9-3x=0$$

Subtract 9 from both sides. Anything subtracted from zero gives its negation.

$$x^{2}-3x=-9$$

Divide -3, the coefficient of the x term, by 2 to get -\frac{3}{2}. Then add the square of -\frac{3}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-3x+\left(-\frac{3}{2}\right)^{2}=-9+\left(-\frac{3}{2}\right)^{2}$$

Square -\frac{3}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-3x+\frac{9}{4}=-9+\frac{9}{4}$$

Add -9 to \frac{9}{4}.

$$x^{2}-3x+\frac{9}{4}=-\frac{27}{4}$$

Factor x^{2}-3x+\frac{9}{4}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{3}{2}\right)^{2}=-\frac{27}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{3}{2}\right)^{2}}=\sqrt{-\frac{27}{4}}$$

Simplify.

$$x-\frac{3}{2}=\frac{3\sqrt{3}i}{2} x-\frac{3}{2}=-\frac{3\sqrt{3}i}{2}$$

Add \frac{3}{2} to both sides of the equation.

$$x=\frac{3+3\sqrt{3}i}{2} x=\frac{-3\sqrt{3}i+3}{2}$$