Solve 64x^3+7=8

$$64 x ^ { 3 } + 7 = 8$$

Solve for x (complex solution)

$$x=\frac{-\sqrt{3}i-1}{8}\approx -0.125-0.216506351i$$

$$x=\frac{1}{4}=0.25$$

$$x=\frac{-1+\sqrt{3}i}{8}\approx -0.125+0.216506351i$$

Solution Steps

Subtract 8 from both sides.

$$64x^{3}+7-8=0$$

Subtract 8 from 7 to get -1.

$$64x^{3}-1=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term -1 and q divides the leading coefficient 64. List all candidates \frac{p}{q}.

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

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.

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

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide 64x^{3}-1 by 4\left(x-\frac{1}{4}\right)=4x-1 to get 16x^{2}+4x+1. Solve the equation where the result equals to 0.

$$16x^{2}+4x+1=0$$

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, 4 for b, and 1 for c in the quadratic formula.

$$x=\frac{-4±\sqrt{4^{2}-4\times 16\times 1}}{2\times 16}$$

Do the calculations.

$$x=\frac{-4±\sqrt{-48}}{32}$$

Solve the equation 16x^{2}+4x+1=0 when ± is plus and when ± is minus.

$$x=\frac{-\sqrt{3}i-1}{8} x=\frac{-1+\sqrt{3}i}{8}$$

List all found solutions.

$$x=\frac{1}{4} x=\frac{-\sqrt{3}i-1}{8} x=\frac{-1+\sqrt{3}i}{8}$$

Solve for x

$$x=\frac{1}{4}=0.25$$

Solution Steps

Subtract 8 from both sides.

$$64x^{3}+7-8=0$$

Subtract 8 from 7 to get -1.

$$64x^{3}-1=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term -1 and q divides the leading coefficient 64. List all candidates \frac{p}{q}.

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

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.

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

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide 64x^{3}-1 by 4\left(x-\frac{1}{4}\right)=4x-1 to get 16x^{2}+4x+1. Solve the equation where the result equals to 0.

$$16x^{2}+4x+1=0$$

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, 4 for b, and 1 for c in the quadratic formula.

$$x=\frac{-4±\sqrt{4^{2}-4\times 16\times 1}}{2\times 16}$$

Do the calculations.

$$x=\frac{-4±\sqrt{-48}}{32}$$

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.

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