Solve (3x-5)^3=0

$$( 3 x - 5 ) ^ { 3 } = 0$$

Solve for x

$$x = \frac{5}{3} = 1\frac{2}{3} \approx 1.666666667$$

Solution Steps

Use binomial theorem \left(a-b\right)^{3}=a^{3}-3a^{2}b+3ab^{2}-b^{3} to expand \left(3x-5\right)^{3}.

$$27x^{3}-135x^{2}+225x-125=0$$

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

$$±\frac{125}{27},±\frac{125}{9},±\frac{125}{3},±125,±\frac{25}{27},±\frac{25}{9},±\frac{25}{3},±25,±\frac{5}{27},±\frac{5}{9},±\frac{5}{3},±5,±\frac{1}{27},±\frac{1}{9},±\frac{1}{3},±1$$

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.

$$x=\frac{5}{3}$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide 27x^{3}-135x^{2}+225x-125 by 3\left(x-\frac{5}{3}\right)=3x-5 to get 9x^{2}-30x+25. Solve the equation where the result equals to 0.

$$9x^{2}-30x+25=0$$

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 9 for a, -30 for b, and 25 for c in the quadratic formula.

$$x=\frac{-\left(-30\right)±\sqrt{\left(-30\right)^{2}-4\times 9\times 25}}{2\times 9}$$

Do the calculations.

$$x=\frac{30±0}{18}$$

Solutions are the same.

$$x=\frac{5}{3}$$