Solve 64x^3+729=0

$$64 x ^ { 3 } + 729 = 0$$

Solve for x (complex solution)

$$x=\frac{-9\sqrt{3}i+9}{8}\approx 1.125-1.948557159i$$

$$x = -\frac{9}{4} = -2\frac{1}{4} = -2.25$$

$$x=\frac{9+9\sqrt{3}i}{8}\approx 1.125+1.948557159i$$

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

$$±\frac{729}{64},±\frac{729}{32},±\frac{729}{16},±\frac{729}{8},±\frac{729}{4},±\frac{729}{2},±729,±\frac{243}{64},±\frac{243}{32},±\frac{243}{16},±\frac{243}{8},±\frac{243}{4},±\frac{243}{2},±243,±\frac{81}{64},±\frac{81}{32},±\frac{81}{16},±\frac{81}{8},±\frac{81}{4},±\frac{81}{2},±81,±\frac{27}{64},±\frac{27}{32},±\frac{27}{16},±\frac{27}{8},±\frac{27}{4},±\frac{27}{2},±27,±\frac{9}{64},±\frac{9}{32},±\frac{9}{16},±\frac{9}{8},±\frac{9}{4},±\frac{9}{2},±9,±\frac{3}{64},±\frac{3}{32},±\frac{3}{16},±\frac{3}{8},±\frac{3}{4},±\frac{3}{2},±3,±\frac{1}{64},±\frac{1}{32},±\frac{1}{16},±\frac{1}{8},±\frac{1}{4},±\frac{1}{2},±1$$

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.

$$x=-\frac{9}{4}$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide 64x^{3}+729 by 4\left(x+\frac{9}{4}\right)=4x+9 to get 16x^{2}-36x+81. Solve the equation where the result equals to 0.

$$16x^{2}-36x+81=0$$

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, -36 for b, and 81 for c in the quadratic formula.

$$x=\frac{-\left(-36\right)±\sqrt{\left(-36\right)^{2}-4\times 16\times 81}}{2\times 16}$$

Do the calculations.

$$x=\frac{36±\sqrt{-3888}}{32}$$

Solve the equation 16x^{2}-36x+81=0 when ± is plus and when ± is minus.

$$x=\frac{-9i\sqrt{3}+9}{8} x=\frac{9+9i\sqrt{3}}{8}$$

List all found solutions.

$$x=-\frac{9}{4} x=\frac{-9i\sqrt{3}+9}{8} x=\frac{9+9i\sqrt{3}}{8}$$

Solve for x

$$x = -\frac{9}{4} = -2\frac{1}{4} = -2.25$$

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

$$±\frac{729}{64},±\frac{729}{32},±\frac{729}{16},±\frac{729}{8},±\frac{729}{4},±\frac{729}{2},±729,±\frac{243}{64},±\frac{243}{32},±\frac{243}{16},±\frac{243}{8},±\frac{243}{4},±\frac{243}{2},±243,±\frac{81}{64},±\frac{81}{32},±\frac{81}{16},±\frac{81}{8},±\frac{81}{4},±\frac{81}{2},±81,±\frac{27}{64},±\frac{27}{32},±\frac{27}{16},±\frac{27}{8},±\frac{27}{4},±\frac{27}{2},±27,±\frac{9}{64},±\frac{9}{32},±\frac{9}{16},±\frac{9}{8},±\frac{9}{4},±\frac{9}{2},±9,±\frac{3}{64},±\frac{3}{32},±\frac{3}{16},±\frac{3}{8},±\frac{3}{4},±\frac{3}{2},±3,±\frac{1}{64},±\frac{1}{32},±\frac{1}{16},±\frac{1}{8},±\frac{1}{4},±\frac{1}{2},±1$$

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.

$$x=-\frac{9}{4}$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide 64x^{3}+729 by 4\left(x+\frac{9}{4}\right)=4x+9 to get 16x^{2}-36x+81. Solve the equation where the result equals to 0.

$$16x^{2}-36x+81=0$$

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, -36 for b, and 81 for c in the quadratic formula.

$$x=\frac{-\left(-36\right)±\sqrt{\left(-36\right)^{2}-4\times 16\times 81}}{2\times 16}$$

Do the calculations.

$$x=\frac{36±\sqrt{-3888}}{32}$$

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.

$$x=-\frac{9}{4}$$