Solve 3left(x-4right)^3=-375

$$3 { \left(x-4 \right) }^{ 3 } = -375$$

Solve for x (complex solution)

$$x=-1$$

$$x=\frac{13+5\sqrt{3}i}{2}\approx 6.5+4.330127019i$$

$$x=\frac{-5\sqrt{3}i+13}{2}\approx 6.5-4.330127019i$$

Solution Steps

Divide both sides by 3.

$$\left(x-4\right)^{3}=\frac{-375}{3}$$

Divide -375 by 3 to get -125.

$$\left(x-4\right)^{3}=-125$$

Use binomial theorem \left(a-b\right)^{3}=a^{3}-3a^{2}b+3ab^{2}-b^{3} to expand \left(x-4\right)^{3}.

$$x^{3}-12x^{2}+48x-64=-125$$

Add 125 to both sides.

$$x^{3}-12x^{2}+48x-64+125=0$$

Add -64 and 125 to get 61.

$$x^{3}-12x^{2}+48x+61=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term 61 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±61,±1$$

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}-12x^{2}+48x+61 by x+1 to get x^{2}-13x+61. Solve the equation where the result equals to 0.

$$x^{2}-13x+61=0$$

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, -13 for b, and 61 for c in the quadratic formula.

$$x=\frac{-\left(-13\right)±\sqrt{\left(-13\right)^{2}-4\times 1\times 61}}{2}$$

Do the calculations.

$$x=\frac{13±\sqrt{-75}}{2}$$

Solve the equation x^{2}-13x+61=0 when ± is plus and when ± is minus.

$$x=\frac{-5i\sqrt{3}+13}{2} x=\frac{13+5i\sqrt{3}}{2}$$

List all found solutions.

$$x=-1 x=\frac{-5i\sqrt{3}+13}{2} x=\frac{13+5i\sqrt{3}}{2}$$

Solve for x

$$x=-1$$

Solution Steps

Divide both sides by 3.

$$\left(x-4\right)^{3}=\frac{-375}{3}$$

Divide -375 by 3 to get -125.

$$\left(x-4\right)^{3}=-125$$

Use binomial theorem \left(a-b\right)^{3}=a^{3}-3a^{2}b+3ab^{2}-b^{3} to expand \left(x-4\right)^{3}.

$$x^{3}-12x^{2}+48x-64=-125$$

Add 125 to both sides.

$$x^{3}-12x^{2}+48x-64+125=0$$

Add -64 and 125 to get 61.

$$x^{3}-12x^{2}+48x+61=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term 61 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±61,±1$$

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}-12x^{2}+48x+61 by x+1 to get x^{2}-13x+61. Solve the equation where the result equals to 0.

$$x^{2}-13x+61=0$$

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, -13 for b, and 61 for c in the quadratic formula.

$$x=\frac{-\left(-13\right)±\sqrt{\left(-13\right)^{2}-4\times 1\times 61}}{2}$$

Do the calculations.

$$x=\frac{13±\sqrt{-75}}{2}$$

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.

$$x=-1$$