Solve 3x^3=375

$$3 x ^ { 3 } = 375$$

Solve for x (complex solution)

$$x=\frac{-5\sqrt{3}i-5}{2}\approx -2.5-4.330127019i$$

$$x=5$$

$$x=\frac{-5+5\sqrt{3}i}{2}\approx -2.5+4.330127019i$$

Solution Steps

Divide both sides by 3.

$$x^{3}=\frac{375}{3}$$

Divide 375 by 3 to get 125.

$$x^{3}=125$$

Subtract 125 from both sides.

$$x^{3}-125=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term -125 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±125,±25,±5,±1$$

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}-125 by x-5 to get x^{2}+5x+25. Solve the equation where the result equals to 0.

$$x^{2}+5x+25=0$$

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, 5 for b, and 25 for c in the quadratic formula.

$$x=\frac{-5±\sqrt{5^{2}-4\times 1\times 25}}{2}$$

Do the calculations.

$$x=\frac{-5±\sqrt{-75}}{2}$$

Solve the equation x^{2}+5x+25=0 when ± is plus and when ± is minus.

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

List all found solutions.

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

Solve for x

$$x=5$$

Solution Steps

Divide both sides by 3.

$$x^{3}=\frac{375}{3}$$

Divide 375 by 3 to get 125.

$$x^{3}=125$$

Subtract 125 from both sides.

$$x^{3}-125=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term -125 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±125,±25,±5,±1$$

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}-125 by x-5 to get x^{2}+5x+25. Solve the equation where the result equals to 0.

$$x^{2}+5x+25=0$$

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, 5 for b, and 25 for c in the quadratic formula.

$$x=\frac{-5±\sqrt{5^{2}-4\times 1\times 25}}{2}$$

Do the calculations.

$$x=\frac{-5±\sqrt{-75}}{2}$$

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.

$$x=5$$