Solve 8=x^3-1/x-1-x^4-1/x-1

$$8= \frac{ { x }^{ 3 } -1 }{ x-1 } - \frac{ { x }^{ 4 } -1 }{ x-1 }$$

Solve for x (complex solution)

$$x=-\sqrt{3}i+1\approx 1-1.732050808i$$

$$x=-2$$

$$x=1+\sqrt{3}i\approx 1+1.732050808i$$

Solution Steps

Variable x cannot be equal to 1 since division by zero is not defined. Multiply both sides of the equation by x-1.

$$8\left(x-1\right)=x^{3}-1-\left(x^{4}-1\right)$$

Use the distributive property to multiply 8 by x-1.

$$8x-8=x^{3}-1-\left(x^{4}-1\right)$$

To find the opposite of x^{4}-1, find the opposite of each term.

$$8x-8=x^{3}-1-x^{4}+1$$

Add -1 and 1 to get 0.

$$8x-8=x^{3}-x^{4}$$

Subtract x^{3} from both sides.

$$8x-8-x^{3}=-x^{4}$$

Add x^{4} to both sides.

$$8x-8-x^{3}+x^{4}=0$$

Rearrange the equation to put it in standard form. Place the terms in order from highest to lowest power.

$$x^{4}-x^{3}+8x-8=0$$

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

$$±8,±4,±2,±1$$

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.

$$x=1$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide x^{4}-x^{3}+8x-8 by x-1 to get x^{3}+8. Solve the equation where the result equals to 0.

$$x^{3}+8=0$$

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

$$±8,±4,±2,±1$$

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.

$$x=-2$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide x^{3}+8 by x+2 to get x^{2}-2x+4. Solve the equation where the result equals to 0.

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

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 4 for c in the quadratic formula.

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

Do the calculations.

$$x=\frac{2±\sqrt{-12}}{2}$$

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

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

Remove the values that the variable cannot be equal to.

$$x=-2$$

List all found solutions.

$$x=1 x=-2 x=-\sqrt{3}i+1 x=1+\sqrt{3}i$$

Variable x cannot be equal to 1.

$$x=1+\sqrt{3}i x=-\sqrt{3}i+1 x=-2$$

Solve for x

$$x=-2$$

Solution Steps

Variable x cannot be equal to 1 since division by zero is not defined. Multiply both sides of the equation by x-1.

$$8\left(x-1\right)=x^{3}-1-\left(x^{4}-1\right)$$

Use the distributive property to multiply 8 by x-1.

$$8x-8=x^{3}-1-\left(x^{4}-1\right)$$

To find the opposite of x^{4}-1, find the opposite of each term.

$$8x-8=x^{3}-1-x^{4}+1$$

Add -1 and 1 to get 0.

$$8x-8=x^{3}-x^{4}$$

Subtract x^{3} from both sides.

$$8x-8-x^{3}=-x^{4}$$

Add x^{4} to both sides.

$$8x-8-x^{3}+x^{4}=0$$

Rearrange the equation to put it in standard form. Place the terms in order from highest to lowest power.

$$x^{4}-x^{3}+8x-8=0$$

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

$$±8,±4,±2,±1$$

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.

$$x=1$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide x^{4}-x^{3}+8x-8 by x-1 to get x^{3}+8. Solve the equation where the result equals to 0.

$$x^{3}+8=0$$

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

$$±8,±4,±2,±1$$

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.

$$x=-2$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide x^{3}+8 by x+2 to get x^{2}-2x+4. Solve the equation where the result equals to 0.

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

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 4 for c in the quadratic formula.

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

Do the calculations.

$$x=\frac{2±\sqrt{-12}}{2}$$

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

$$x\in \emptyset$$

Remove the values that the variable cannot be equal to.

$$x=-2$$

List all found solutions.

$$x=1 x=-2$$

Variable x cannot be equal to 1.

$$x=-2$$