Solve 8(x-3)^3=64

$$8 ( x - 3 ) ^ { 3 } = 64$$

Solve for x (complex solution)

$$x=5$$

$$x=2+\sqrt{3}i\approx 2+1.732050808i$$

$$x=-\sqrt{3}i+2\approx 2-1.732050808i$$

Solution Steps

Divide both sides by 8.

$$\left(x-3\right)^{3}=\frac{64}{8}$$

Divide 64 by 8 to get 8.

$$\left(x-3\right)^{3}=8$$

Use binomial theorem \left(a-b\right)^{3}=a^{3}-3a^{2}b+3ab^{2}-b^{3} to expand \left(x-3\right)^{3}.

$$x^{3}-9x^{2}+27x-27=8$$

Subtract 8 from both sides.

$$x^{3}-9x^{2}+27x-27-8=0$$

Subtract 8 from -27 to get -35.

$$x^{3}-9x^{2}+27x-35=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term -35 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±35,±7,±5,±1$$

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.

$$x=5$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide x^{3}-9x^{2}+27x-35 by x-5 to get x^{2}-4x+7. Solve the equation where the result equals to 0.

$$x^{2}-4x+7=0$$

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, -4 for b, and 7 for c in the quadratic formula.

$$x=\frac{-\left(-4\right)±\sqrt{\left(-4\right)^{2}-4\times 1\times 7}}{2}$$

Do the calculations.

$$x=\frac{4±\sqrt{-12}}{2}$$

Solve the equation x^{2}-4x+7=0 when ± is plus and when ± is minus.

$$x=-\sqrt{3}i+2 x=2+\sqrt{3}i$$

List all found solutions.

$$x=5 x=-\sqrt{3}i+2 x=2+\sqrt{3}i$$

Solve for x

$$x=5$$

Solution Steps

Divide both sides by 8.

$$\left(x-3\right)^{3}=\frac{64}{8}$$

Divide 64 by 8 to get 8.

$$\left(x-3\right)^{3}=8$$

Use binomial theorem \left(a-b\right)^{3}=a^{3}-3a^{2}b+3ab^{2}-b^{3} to expand \left(x-3\right)^{3}.

$$x^{3}-9x^{2}+27x-27=8$$

Subtract 8 from both sides.

$$x^{3}-9x^{2}+27x-27-8=0$$

Subtract 8 from -27 to get -35.

$$x^{3}-9x^{2}+27x-35=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term -35 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±35,±7,±5,±1$$

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.

$$x=5$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide x^{3}-9x^{2}+27x-35 by x-5 to get x^{2}-4x+7. Solve the equation where the result equals to 0.

$$x^{2}-4x+7=0$$

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, -4 for b, and 7 for c in the quadratic formula.

$$x=\frac{-\left(-4\right)±\sqrt{\left(-4\right)^{2}-4\times 1\times 7}}{2}$$

Do the calculations.

$$x=\frac{4±\sqrt{-12}}{2}$$

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.

$$x=5$$