Solve 8(x-7)^3+27=0

$$8 ( x - 7 ) ^ { 3 } + 27 = 0$$

Solve for x (complex solution)

$$x = \frac{11}{2} = 5\frac{1}{2} = 5.5$$

$$x=\frac{31+3\sqrt{3}i}{4}\approx 7.75+1.299038106i$$

$$x=\frac{-3\sqrt{3}i+31}{4}\approx 7.75-1.299038106i$$

Solution Steps

Expand the expression.

$$8x^{3}-168x^{2}+1176x-2717=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term -2717 and q divides the leading coefficient 8. List all candidates \frac{p}{q}.

$$±\frac{2717}{8},±\frac{2717}{4},±\frac{2717}{2},±2717,±\frac{247}{8},±\frac{247}{4},±\frac{247}{2},±247,±\frac{209}{8},±\frac{209}{4},±\frac{209}{2},±209,±\frac{143}{8},±\frac{143}{4},±\frac{143}{2},±143,±\frac{19}{8},±\frac{19}{4},±\frac{19}{2},±19,±\frac{13}{8},±\frac{13}{4},±\frac{13}{2},±13,±\frac{11}{8},±\frac{11}{4},±\frac{11}{2},±11,±\frac{1}{8},±\frac{1}{4},±\frac{1}{2},±1$$

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.

$$x=\frac{11}{2}$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide 8x^{3}-168x^{2}+1176x-2717 by 2\left(x-\frac{11}{2}\right)=2x-11 to get 4x^{2}-62x+247. Solve the equation where the result equals to 0.

$$4x^{2}-62x+247=0$$

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, -62 for b, and 247 for c in the quadratic formula.

$$x=\frac{-\left(-62\right)±\sqrt{\left(-62\right)^{2}-4\times 4\times 247}}{2\times 4}$$

Do the calculations.

$$x=\frac{62±\sqrt{-108}}{8}$$

Solve the equation 4x^{2}-62x+247=0 when ± is plus and when ± is minus.

$$x=\frac{-3i\sqrt{3}+31}{4} x=\frac{31+3i\sqrt{3}}{4}$$

List all found solutions.

$$x=\frac{11}{2} x=\frac{-3i\sqrt{3}+31}{4} x=\frac{31+3i\sqrt{3}}{4}$$

Solve for x

$$x = \frac{11}{2} = 5\frac{1}{2} = 5.5$$

Solution Steps

Expand the expression.

$$8x^{3}-168x^{2}+1176x-2717=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term -2717 and q divides the leading coefficient 8. List all candidates \frac{p}{q}.

$$±\frac{2717}{8},±\frac{2717}{4},±\frac{2717}{2},±2717,±\frac{247}{8},±\frac{247}{4},±\frac{247}{2},±247,±\frac{209}{8},±\frac{209}{4},±\frac{209}{2},±209,±\frac{143}{8},±\frac{143}{4},±\frac{143}{2},±143,±\frac{19}{8},±\frac{19}{4},±\frac{19}{2},±19,±\frac{13}{8},±\frac{13}{4},±\frac{13}{2},±13,±\frac{11}{8},±\frac{11}{4},±\frac{11}{2},±11,±\frac{1}{8},±\frac{1}{4},±\frac{1}{2},±1$$

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.

$$x=\frac{11}{2}$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide 8x^{3}-168x^{2}+1176x-2717 by 2\left(x-\frac{11}{2}\right)=2x-11 to get 4x^{2}-62x+247. Solve the equation where the result equals to 0.

$$4x^{2}-62x+247=0$$

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, -62 for b, and 247 for c in the quadratic formula.

$$x=\frac{-\left(-62\right)±\sqrt{\left(-62\right)^{2}-4\times 4\times 247}}{2\times 4}$$

Do the calculations.

$$x=\frac{62±\sqrt{-108}}{8}$$

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.

$$x=\frac{11}{2}$$