Solve 64x^3-27=0

$$64 { x }^{ 3 } -27=0$$

Solve for x (complex solution)

$$x=\frac{-3\sqrt{3}i-3}{8}\approx -0.375-0.649519053i$$

$$x=\frac{3}{4}=0.75$$

$$x=\frac{-3+3\sqrt{3}i}{8}\approx -0.375+0.649519053i$$

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 -27 and q divides the leading coefficient 64. List all candidates \frac{p}{q}.

$$±\frac{27}{64},±\frac{27}{32},±\frac{27}{16},±\frac{27}{8},±\frac{27}{4},±\frac{27}{2},±27,±\frac{9}{64},±\frac{9}{32},±\frac{9}{16},±\frac{9}{8},±\frac{9}{4},±\frac{9}{2},±9,±\frac{3}{64},±\frac{3}{32},±\frac{3}{16},±\frac{3}{8},±\frac{3}{4},±\frac{3}{2},±3,±\frac{1}{64},±\frac{1}{32},±\frac{1}{16},±\frac{1}{8},±\frac{1}{4},±\frac{1}{2},±1$$

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.

$$x=\frac{3}{4}$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide 64x^{3}-27 by 4\left(x-\frac{3}{4}\right)=4x-3 to get 16x^{2}+12x+9. Solve the equation where the result equals to 0.

$$16x^{2}+12x+9=0$$

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, 12 for b, and 9 for c in the quadratic formula.

$$x=\frac{-12±\sqrt{12^{2}-4\times 16\times 9}}{2\times 16}$$

Do the calculations.

$$x=\frac{-12±\sqrt{-432}}{32}$$

Solve the equation 16x^{2}+12x+9=0 when ± is plus and when ± is minus.

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

List all found solutions.

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

Solve for x

$$x=\frac{3}{4}=0.75$$

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 -27 and q divides the leading coefficient 64. List all candidates \frac{p}{q}.

$$±\frac{27}{64},±\frac{27}{32},±\frac{27}{16},±\frac{27}{8},±\frac{27}{4},±\frac{27}{2},±27,±\frac{9}{64},±\frac{9}{32},±\frac{9}{16},±\frac{9}{8},±\frac{9}{4},±\frac{9}{2},±9,±\frac{3}{64},±\frac{3}{32},±\frac{3}{16},±\frac{3}{8},±\frac{3}{4},±\frac{3}{2},±3,±\frac{1}{64},±\frac{1}{32},±\frac{1}{16},±\frac{1}{8},±\frac{1}{4},±\frac{1}{2},±1$$

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.

$$x=\frac{3}{4}$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide 64x^{3}-27 by 4\left(x-\frac{3}{4}\right)=4x-3 to get 16x^{2}+12x+9. Solve the equation where the result equals to 0.

$$16x^{2}+12x+9=0$$

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, 12 for b, and 9 for c in the quadratic formula.

$$x=\frac{-12±\sqrt{12^{2}-4\times 16\times 9}}{2\times 16}$$

Do the calculations.

$$x=\frac{-12±\sqrt{-432}}{32}$$

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.

$$x=\frac{3}{4}$$