Solve (x/2)^3=125

$$( \frac { x } { 2 } ) ^ { 3 } = 125$$

Solve for x

$$x=10$$

Solution Steps

To raise \frac{x}{2} to a power, raise both numerator and denominator to the power and then divide.

$$\frac{x^{3}}{2^{3}}=125$$

Calculate 2 to the power of 3 and get 8.

$$\frac{x^{3}}{8}=125$$

Subtract 125 from both sides.

$$\frac{x^{3}}{8}-125=0$$

Multiply both sides of the equation by 8.

$$x^{3}-1000=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term -1000 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±1000,±500,±250,±200,±125,±100,±50,±40,±25,±20,±10,±8,±5,±4,±2,±1$$

Find one such root by trying out all the integer values, starting from the smallest by absolute value. If no integer roots are found, try out fractions.

$$x=10$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide x^{3}-1000 by x-10 to get x^{2}+10x+100. Solve the equation where the result equals to 0.

$$x^{2}+10x+100=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. Substitute 1 for a, 10 for b, and 100 for c in the quadratic formula.

$$x=\frac{-10±\sqrt{10^{2}-4\times 1\times 100}}{2}$$

Do the calculations.

$$x=\frac{-10±\sqrt{-300}}{2}$$

Since the square root of a negative number is not defined in the real field, there are no solutions.

$$x\in \emptyset$$

List all found solutions.

$$x=10$$