Solve 64x^3-125=0

$$64 x ^ { 3 } - 125 = 0$$

Solve for x (complex solution)

$$x=\frac{-5\sqrt{3}i-5}{8}\approx -0.625-1.082531755i$$

$$x = \frac{5}{4} = 1\frac{1}{4} = 1.25$$

$$x=\frac{-5+5\sqrt{3}i}{8}\approx -0.625+1.082531755i$$

Solution Steps

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term -125 and q divides the leading coefficient 64. List all candidates \frac{p}{q}.

$$±\frac{125}{64},±\frac{125}{32},±\frac{125}{16},±\frac{125}{8},±\frac{125}{4},±\frac{125}{2},±125,±\frac{25}{64},±\frac{25}{32},±\frac{25}{16},±\frac{25}{8},±\frac{25}{4},±\frac{25}{2},±25,±\frac{5}{64},±\frac{5}{32},±\frac{5}{16},±\frac{5}{8},±\frac{5}{4},±\frac{5}{2},±5,±\frac{1}{64},±\frac{1}{32},±\frac{1}{16},±\frac{1}{8},±\frac{1}{4},±\frac{1}{2},±1$$

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.

$$x=\frac{5}{4}$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide 64x^{3}-125 by 4\left(x-\frac{5}{4}\right)=4x-5 to get 16x^{2}+20x+25. Solve the equation where the result equals to 0.

$$16x^{2}+20x+25=0$$

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 16 for a, 20 for b, and 25 for c in the quadratic formula.

$$x=\frac{-20±\sqrt{20^{2}-4\times 16\times 25}}{2\times 16}$$

Do the calculations.

$$x=\frac{-20±\sqrt{-1200}}{32}$$

Solve the equation 16x^{2}+20x+25=0 when ± is plus and when ± is minus.

$$x=\frac{-5i\sqrt{3}-5}{8} x=\frac{-5+5i\sqrt{3}}{8}$$

List all found solutions.

$$x=\frac{5}{4} x=\frac{-5i\sqrt{3}-5}{8} x=\frac{-5+5i\sqrt{3}}{8}$$

Solve for x

$$x = \frac{5}{4} = 1\frac{1}{4} = 1.25$$

Solution Steps

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term -125 and q divides the leading coefficient 64. List all candidates \frac{p}{q}.

$$±\frac{125}{64},±\frac{125}{32},±\frac{125}{16},±\frac{125}{8},±\frac{125}{4},±\frac{125}{2},±125,±\frac{25}{64},±\frac{25}{32},±\frac{25}{16},±\frac{25}{8},±\frac{25}{4},±\frac{25}{2},±25,±\frac{5}{64},±\frac{5}{32},±\frac{5}{16},±\frac{5}{8},±\frac{5}{4},±\frac{5}{2},±5,±\frac{1}{64},±\frac{1}{32},±\frac{1}{16},±\frac{1}{8},±\frac{1}{4},±\frac{1}{2},±1$$

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.

$$x=\frac{5}{4}$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide 64x^{3}-125 by 4\left(x-\frac{5}{4}\right)=4x-5 to get 16x^{2}+20x+25. Solve the equation where the result equals to 0.

$$16x^{2}+20x+25=0$$

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 16 for a, 20 for b, and 25 for c in the quadratic formula.

$$x=\frac{-20±\sqrt{20^{2}-4\times 16\times 25}}{2\times 16}$$

Do the calculations.

$$x=\frac{-20±\sqrt{-1200}}{32}$$

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.

$$x=\frac{5}{4}$$