Solve 1/12c^3=2/3

$$\frac { 1 } { 12 c ^ { 3 } } = \frac { 2 } { 3 }$$

Solve for c

$$c=\frac{1}{2}=0.5$$

Solution Steps

Variable c cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by 12c^{3}, the least common multiple of 12c^{3},3.

$$1=8c^{3}$$

Swap sides so that all variable terms are on the left hand side.

$$8c^{3}=1$$

Subtract 1 from both sides.

$$8c^{3}-1=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term -1 and q divides the leading coefficient 8. List all candidates \frac{p}{q}.

$$±\frac{1}{8},±\frac{1}{4},±\frac{1}{2},±1$$

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.

$$c=\frac{1}{2}$$

By Factor theorem, c-k is a factor of the polynomial for each root k. Divide 8c^{3}-1 by 2\left(c-\frac{1}{2}\right)=2c-1 to get 4c^{2}+2c+1. Solve the equation where the result equals to 0.

$$4c^{2}+2c+1=0$$

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 4 for a, 2 for b, and 1 for c in the quadratic formula.

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

Do the calculations.

$$c=\frac{-2±\sqrt{-12}}{8}$$

Since the square root of a negative number is not defined in the real field, there are no solutions.

$$c\in \emptyset$$

List all found solutions.

$$c=\frac{1}{2}$$