Solve (x-3)^3=-64

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

Solve for x (complex solution)

$$x=-1$$

$$x=5+2\sqrt{3}i\approx 5+3.464101615i$$

$$x=-2\sqrt{3}i+5\approx 5-3.464101615i$$

Solution Steps

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

$$x^{3}-9x^{2}+27x-27=-64$$

Add 64 to both sides.

$$x^{3}-9x^{2}+27x-27+64=0$$

Add -27 and 64 to get 37.

$$x^{3}-9x^{2}+27x+37=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term 37 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±37,±1$$

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.

$$x=-1$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide x^{3}-9x^{2}+27x+37 by x+1 to get x^{2}-10x+37. Solve the equation where the result equals to 0.

$$x^{2}-10x+37=0$$

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, -10 for b, and 37 for c in the quadratic formula.

$$x=\frac{-\left(-10\right)±\sqrt{\left(-10\right)^{2}-4\times 1\times 37}}{2}$$

Do the calculations.

$$x=\frac{10±\sqrt{-48}}{2}$$

Solve the equation x^{2}-10x+37=0 when ± is plus and when ± is minus.

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

List all found solutions.

$$x=-1 x=-2i\sqrt{3}+5 x=5+2i\sqrt{3}$$

Solve for x

$$x=-1$$

Solution Steps

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

$$x^{3}-9x^{2}+27x-27=-64$$

Add 64 to both sides.

$$x^{3}-9x^{2}+27x-27+64=0$$

Add -27 and 64 to get 37.

$$x^{3}-9x^{2}+27x+37=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term 37 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±37,±1$$

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.

$$x=-1$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide x^{3}-9x^{2}+27x+37 by x+1 to get x^{2}-10x+37. Solve the equation where the result equals to 0.

$$x^{2}-10x+37=0$$

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, -10 for b, and 37 for c in the quadratic formula.

$$x=\frac{-\left(-10\right)±\sqrt{\left(-10\right)^{2}-4\times 1\times 37}}{2}$$

Do the calculations.

$$x=\frac{10±\sqrt{-48}}{2}$$

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.

$$x=-1$$