Solve 27left(2x-1right)^3+2=66

$$27 { \left(2x-1 \right) }^{ 3 } +2=66$$

Solve for x (complex solution)

$$x = \frac{7}{6} = 1\frac{1}{6} \approx 1.166666667$$

$$x=\frac{\sqrt{3}i}{3}+\frac{1}{6}\approx 0.166666667+0.577350269i$$

$$x=-\frac{\sqrt{3}i}{3}+\frac{1}{6}\approx 0.166666667-0.577350269i$$

Solution Steps

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

$$27\left(8x^{3}-12x^{2}+6x-1\right)+2=66$$

Use the distributive property to multiply 27 by 8x^{3}-12x^{2}+6x-1.

$$216x^{3}-324x^{2}+162x-27+2=66$$

Add -27 and 2 to get -25.

$$216x^{3}-324x^{2}+162x-25=66$$

Subtract 66 from both sides.

$$216x^{3}-324x^{2}+162x-25-66=0$$

Subtract 66 from -25 to get -91.

$$216x^{3}-324x^{2}+162x-91=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term -91 and q divides the leading coefficient 216. List all candidates \frac{p}{q}.

$$±\frac{91}{216},±\frac{91}{108},±\frac{91}{72},±\frac{91}{54},±\frac{91}{36},±\frac{91}{27},±\frac{91}{24},±\frac{91}{18},±\frac{91}{12},±\frac{91}{9},±\frac{91}{8},±\frac{91}{6},±\frac{91}{4},±\frac{91}{3},±\frac{91}{2},±91,±\frac{13}{216},±\frac{13}{108},±\frac{13}{72},±\frac{13}{54},±\frac{13}{36},±\frac{13}{27},±\frac{13}{24},±\frac{13}{18},±\frac{13}{12},±\frac{13}{9},±\frac{13}{8},±\frac{13}{6},±\frac{13}{4},±\frac{13}{3},±\frac{13}{2},±13,±\frac{7}{216},±\frac{7}{108},±\frac{7}{72},±\frac{7}{54},±\frac{7}{36},±\frac{7}{27},±\frac{7}{24},±\frac{7}{18},±\frac{7}{12},±\frac{7}{9},±\frac{7}{8},±\frac{7}{6},±\frac{7}{4},±\frac{7}{3},±\frac{7}{2},±7,±\frac{1}{216},±\frac{1}{108},±\frac{1}{72},±\frac{1}{54},±\frac{1}{36},±\frac{1}{27},±\frac{1}{24},±\frac{1}{18},±\frac{1}{12},±\frac{1}{9},±\frac{1}{8},±\frac{1}{6},±\frac{1}{4},±\frac{1}{3},±\frac{1}{2},±1$$

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.

$$x=\frac{7}{6}$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide 216x^{3}-324x^{2}+162x-91 by 6\left(x-\frac{7}{6}\right)=6x-7 to get 36x^{2}-12x+13. Solve the equation where the result equals to 0.

$$36x^{2}-12x+13=0$$

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 36 for a, -12 for b, and 13 for c in the quadratic formula.

$$x=\frac{-\left(-12\right)±\sqrt{\left(-12\right)^{2}-4\times 36\times 13}}{2\times 36}$$

Do the calculations.

$$x=\frac{12±\sqrt{-1728}}{72}$$

Solve the equation 36x^{2}-12x+13=0 when ± is plus and when ± is minus.

$$x=-\frac{\sqrt{3}i}{3}+\frac{1}{6} x=\frac{\sqrt{3}i}{3}+\frac{1}{6}$$

List all found solutions.

$$x=\frac{7}{6} x=-\frac{\sqrt{3}i}{3}+\frac{1}{6} x=\frac{\sqrt{3}i}{3}+\frac{1}{6}$$

Solve for x

$$x = \frac{7}{6} = 1\frac{1}{6} \approx 1.166666667$$

Solution Steps

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

$$27\left(8x^{3}-12x^{2}+6x-1\right)+2=66$$

Use the distributive property to multiply 27 by 8x^{3}-12x^{2}+6x-1.

$$216x^{3}-324x^{2}+162x-27+2=66$$

Add -27 and 2 to get -25.

$$216x^{3}-324x^{2}+162x-25=66$$

Subtract 66 from both sides.

$$216x^{3}-324x^{2}+162x-25-66=0$$

Subtract 66 from -25 to get -91.

$$216x^{3}-324x^{2}+162x-91=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term -91 and q divides the leading coefficient 216. List all candidates \frac{p}{q}.

$$±\frac{91}{216},±\frac{91}{108},±\frac{91}{72},±\frac{91}{54},±\frac{91}{36},±\frac{91}{27},±\frac{91}{24},±\frac{91}{18},±\frac{91}{12},±\frac{91}{9},±\frac{91}{8},±\frac{91}{6},±\frac{91}{4},±\frac{91}{3},±\frac{91}{2},±91,±\frac{13}{216},±\frac{13}{108},±\frac{13}{72},±\frac{13}{54},±\frac{13}{36},±\frac{13}{27},±\frac{13}{24},±\frac{13}{18},±\frac{13}{12},±\frac{13}{9},±\frac{13}{8},±\frac{13}{6},±\frac{13}{4},±\frac{13}{3},±\frac{13}{2},±13,±\frac{7}{216},±\frac{7}{108},±\frac{7}{72},±\frac{7}{54},±\frac{7}{36},±\frac{7}{27},±\frac{7}{24},±\frac{7}{18},±\frac{7}{12},±\frac{7}{9},±\frac{7}{8},±\frac{7}{6},±\frac{7}{4},±\frac{7}{3},±\frac{7}{2},±7,±\frac{1}{216},±\frac{1}{108},±\frac{1}{72},±\frac{1}{54},±\frac{1}{36},±\frac{1}{27},±\frac{1}{24},±\frac{1}{18},±\frac{1}{12},±\frac{1}{9},±\frac{1}{8},±\frac{1}{6},±\frac{1}{4},±\frac{1}{3},±\frac{1}{2},±1$$

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.

$$x=\frac{7}{6}$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide 216x^{3}-324x^{2}+162x-91 by 6\left(x-\frac{7}{6}\right)=6x-7 to get 36x^{2}-12x+13. Solve the equation where the result equals to 0.

$$36x^{2}-12x+13=0$$

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 36 for a, -12 for b, and 13 for c in the quadratic formula.

$$x=\frac{-\left(-12\right)±\sqrt{\left(-12\right)^{2}-4\times 36\times 13}}{2\times 36}$$

Do the calculations.

$$x=\frac{12±\sqrt{-1728}}{72}$$

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.

$$x=\frac{7}{6}$$