Solve =4(c-1)^3=0

$$= 4 ( c - 1 ) ^ { 3 } = 0$$

Solve for c

$$c=1$$

Solution Steps

Divide both sides by 4. Zero divided by any non-zero number gives zero.

$$\left(c-1\right)^{3}=0$$

Use binomial theorem \left(a-b\right)^{3}=a^{3}-3a^{2}b+3ab^{2}-b^{3} to expand \left(c-1\right)^{3}.

$$c^{3}-3c^{2}+3c-1=0$$

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 1. List all candidates \frac{p}{q}.

$$±1$$

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.

$$c=1$$

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

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

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

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

Do the calculations.

$$c=\frac{2±0}{2}$$

Solutions are the same.

$$c=1$$