Solve 3x^3=192

$$3 { x }^{ 3 } =192$$

Solve for x (complex solution)

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

$$x=4$$

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

Solution Steps

Divide both sides by 3.

$$x^{3}=\frac{192}{3}$$

Divide 192 by 3 to get 64.

$$x^{3}=64$$

Subtract 64 from both sides.

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

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

$$±64,±32,±16,±8,±4,±2,±1$$

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.

$$x=4$$

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

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

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 16 for c in the quadratic formula.

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

Do the calculations.

$$x=\frac{-4±\sqrt{-48}}{2}$$

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

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

List all found solutions.

$$x=4 x=-2i\sqrt{3}-2 x=-2+2i\sqrt{3}$$

Solve for x

$$x=4$$

Solution Steps

Divide both sides by 3.

$$x^{3}=\frac{192}{3}$$

Divide 192 by 3 to get 64.

$$x^{3}=64$$

Subtract 64 from both sides.

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

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

$$±64,±32,±16,±8,±4,±2,±1$$

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.

$$x=4$$

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

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

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 16 for c in the quadratic formula.

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

Do the calculations.

$$x=\frac{-4±\sqrt{-48}}{2}$$

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.

$$x=4$$