Solve 6/x^3+1-1-x/x^2-x+1=5/x+1

$$\frac{ 6 }{ { x }^{ 3 } +1 } - \frac{ 1-x }{ { x }^{ 2 } -x+1 } = \frac{ 5 }{ x+1 }$$

Solve for x

$$x = \frac{5}{4} = 1\frac{1}{4} = 1.25$$

$$x=0$$

Steps Using Factoring

Variable x cannot be equal to -1 since division by zero is not defined. Multiply both sides of the equation by \left(x+1\right)\left(x^{2}-x+1\right), the least common multiple of x^{3}+1,x^{2}-x+1,x+1.

$$6-\left(x+1\right)\left(1-x\right)=\left(x^{2}-x+1\right)\times 5$$

Consider \left(x+1\right)\left(1-x\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}. Square 1.

$$6-\left(1-x^{2}\right)=\left(x^{2}-x+1\right)\times 5$$

To find the opposite of 1-x^{2}, find the opposite of each term.

$$6-1+x^{2}=\left(x^{2}-x+1\right)\times 5$$

Subtract 1 from 6 to get 5.

$$5+x^{2}=\left(x^{2}-x+1\right)\times 5$$

Use the distributive property to multiply x^{2}-x+1 by 5.

$$5+x^{2}=5x^{2}-5x+5$$

Subtract 5x^{2} from both sides.

$$5+x^{2}-5x^{2}=-5x+5$$

Combine x^{2} and -5x^{2} to get -4x^{2}.

$$5-4x^{2}=-5x+5$$

Add 5x to both sides.

$$5-4x^{2}+5x=5$$

Subtract 5 from both sides.

$$5-4x^{2}+5x-5=0$$

Subtract 5 from 5 to get 0.

$$-4x^{2}+5x=0$$

Factor out x.

$$x\left(-4x+5\right)=0$$

To find equation solutions, solve x=0 and -4x+5=0.

$$x=0 x=\frac{5}{4}$$

Steps Using the Quadratic Formula

Variable x cannot be equal to -1 since division by zero is not defined. Multiply both sides of the equation by \left(x+1\right)\left(x^{2}-x+1\right), the least common multiple of x^{3}+1,x^{2}-x+1,x+1.

$$6-\left(x+1\right)\left(1-x\right)=\left(x^{2}-x+1\right)\times 5$$

Consider \left(x+1\right)\left(1-x\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}. Square 1.

$$6-\left(1-x^{2}\right)=\left(x^{2}-x+1\right)\times 5$$

To find the opposite of 1-x^{2}, find the opposite of each term.

$$6-1+x^{2}=\left(x^{2}-x+1\right)\times 5$$

Subtract 1 from 6 to get 5.

$$5+x^{2}=\left(x^{2}-x+1\right)\times 5$$

Use the distributive property to multiply x^{2}-x+1 by 5.

$$5+x^{2}=5x^{2}-5x+5$$

Subtract 5x^{2} from both sides.

$$5+x^{2}-5x^{2}=-5x+5$$

Combine x^{2} and -5x^{2} to get -4x^{2}.

$$5-4x^{2}=-5x+5$$

Add 5x to both sides.

$$5-4x^{2}+5x=5$$

Subtract 5 from both sides.

$$5-4x^{2}+5x-5=0$$

Subtract 5 from 5 to get 0.

$$-4x^{2}+5x=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -4 for a, 5 for b, and 0 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-5±\sqrt{5^{2}}}{2\left(-4\right)}$$

Take the square root of 5^{2}.

$$x=\frac{-5±5}{2\left(-4\right)}$$

Multiply 2 times -4.

$$x=\frac{-5±5}{-8}$$

Now solve the equation x=\frac{-5±5}{-8} when ± is plus. Add -5 to 5.

$$x=\frac{0}{-8}$$

Divide 0 by -8.

$$x=0$$

Now solve the equation x=\frac{-5±5}{-8} when ± is minus. Subtract 5 from -5.

$$x=-\frac{10}{-8}$$

Reduce the fraction \frac{-10}{-8} to lowest terms by extracting and canceling out 2.

$$x=\frac{5}{4}$$

The equation is now solved.

$$x=0 x=\frac{5}{4}$$

Steps for Completing the Square

Variable x cannot be equal to -1 since division by zero is not defined. Multiply both sides of the equation by \left(x+1\right)\left(x^{2}-x+1\right), the least common multiple of x^{3}+1,x^{2}-x+1,x+1.

$$6-\left(x+1\right)\left(1-x\right)=\left(x^{2}-x+1\right)\times 5$$

Consider \left(x+1\right)\left(1-x\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}. Square 1.

$$6-\left(1-x^{2}\right)=\left(x^{2}-x+1\right)\times 5$$

To find the opposite of 1-x^{2}, find the opposite of each term.

$$6-1+x^{2}=\left(x^{2}-x+1\right)\times 5$$

Subtract 1 from 6 to get 5.

$$5+x^{2}=\left(x^{2}-x+1\right)\times 5$$

Use the distributive property to multiply x^{2}-x+1 by 5.

$$5+x^{2}=5x^{2}-5x+5$$

Subtract 5x^{2} from both sides.

$$5+x^{2}-5x^{2}=-5x+5$$

Combine x^{2} and -5x^{2} to get -4x^{2}.

$$5-4x^{2}=-5x+5$$

Add 5x to both sides.

$$5-4x^{2}+5x=5$$

Subtract 5 from both sides.

$$-4x^{2}+5x=5-5$$

Subtract 5 from 5 to get 0.

$$-4x^{2}+5x=0$$

Divide both sides by -4.

$$\frac{-4x^{2}+5x}{-4}=\frac{0}{-4}$$

Dividing by -4 undoes the multiplication by -4.

$$x^{2}+\frac{5}{-4}x=\frac{0}{-4}$$

Divide 5 by -4.

$$x^{2}-\frac{5}{4}x=\frac{0}{-4}$$

Divide 0 by -4.

$$x^{2}-\frac{5}{4}x=0$$

Divide -\frac{5}{4}, the coefficient of the x term, by 2 to get -\frac{5}{8}. Then add the square of -\frac{5}{8} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{5}{4}x+\left(-\frac{5}{8}\right)^{2}=\left(-\frac{5}{8}\right)^{2}$$

Square -\frac{5}{8} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{5}{4}x+\frac{25}{64}=\frac{25}{64}$$

Factor x^{2}-\frac{5}{4}x+\frac{25}{64}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{5}{8}\right)^{2}=\frac{25}{64}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{5}{8}\right)^{2}}=\sqrt{\frac{25}{64}}$$

Simplify.

$$x-\frac{5}{8}=\frac{5}{8} x-\frac{5}{8}=-\frac{5}{8}$$

Add \frac{5}{8} to both sides of the equation.

$$x=\frac{5}{4} x=0$$