Solve x^3-1/(x-1)^2-x^3+1/(x+1)^2=6

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

Solve for x

$$x=2$$

$$x=-2$$

Steps Using the Quadratic Formula

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

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

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

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

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

$$x^{5}-x^{2}+2x^{4}-2x+x^{3}-1-\left(x-1\right)^{2}\left(x^{3}+1\right)=6\left(x-1\right)^{2}\left(x+1\right)^{2}$$

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

$$x^{5}-x^{2}+2x^{4}-2x+x^{3}-1-\left(x^{2}-2x+1\right)\left(x^{3}+1\right)=6\left(x-1\right)^{2}\left(x+1\right)^{2}$$

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

$$x^{5}-x^{2}+2x^{4}-2x+x^{3}-1-\left(x^{5}+x^{2}-2x^{4}-2x+x^{3}+1\right)=6\left(x-1\right)^{2}\left(x+1\right)^{2}$$

To find the opposite of x^{5}+x^{2}-2x^{4}-2x+x^{3}+1, find the opposite of each term.

$$x^{5}-x^{2}+2x^{4}-2x+x^{3}-1-x^{5}-x^{2}+2x^{4}+2x-x^{3}-1=6\left(x-1\right)^{2}\left(x+1\right)^{2}$$

Combine x^{5} and -x^{5} to get 0.

$$-x^{2}+2x^{4}-2x+x^{3}-1-x^{2}+2x^{4}+2x-x^{3}-1=6\left(x-1\right)^{2}\left(x+1\right)^{2}$$

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

$$-2x^{2}+2x^{4}-2x+x^{3}-1+2x^{4}+2x-x^{3}-1=6\left(x-1\right)^{2}\left(x+1\right)^{2}$$

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

$$-2x^{2}+4x^{4}-2x+x^{3}-1+2x-x^{3}-1=6\left(x-1\right)^{2}\left(x+1\right)^{2}$$

Combine -2x and 2x to get 0.

$$-2x^{2}+4x^{4}+x^{3}-1-x^{3}-1=6\left(x-1\right)^{2}\left(x+1\right)^{2}$$

Combine x^{3} and -x^{3} to get 0.

$$-2x^{2}+4x^{4}-1-1=6\left(x-1\right)^{2}\left(x+1\right)^{2}$$

Subtract 1 from -1 to get -2.

$$-2x^{2}+4x^{4}-2=6\left(x-1\right)^{2}\left(x+1\right)^{2}$$

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

$$-2x^{2}+4x^{4}-2=6\left(x^{2}-2x+1\right)\left(x+1\right)^{2}$$

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

$$-2x^{2}+4x^{4}-2=6\left(x^{2}-2x+1\right)\left(x^{2}+2x+1\right)$$

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

$$-2x^{2}+4x^{4}-2=\left(6x^{2}-12x+6\right)\left(x^{2}+2x+1\right)$$

Use the distributive property to multiply 6x^{2}-12x+6 by x^{2}+2x+1 and combine like terms.

$$-2x^{2}+4x^{4}-2=6x^{4}-12x^{2}+6$$

Subtract 6x^{4} from both sides.

$$-2x^{2}+4x^{4}-2-6x^{4}=-12x^{2}+6$$

Combine 4x^{4} and -6x^{4} to get -2x^{4}.

$$-2x^{2}-2x^{4}-2=-12x^{2}+6$$

Add 12x^{2} to both sides.

$$-2x^{2}-2x^{4}-2+12x^{2}=6$$

Combine -2x^{2} and 12x^{2} to get 10x^{2}.

$$10x^{2}-2x^{4}-2=6$$

Subtract 6 from both sides.

$$10x^{2}-2x^{4}-2-6=0$$

Subtract 6 from -2 to get -8.

$$10x^{2}-2x^{4}-8=0$$

Substitute t for x^{2}.

$$-2t^{2}+10t-8=0$$

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 -2 for a, 10 for b, and -8 for c in the quadratic formula.

$$t=\frac{-10±\sqrt{10^{2}-4\left(-2\right)\left(-8\right)}}{-2\times 2}$$

Do the calculations.

$$t=\frac{-10±6}{-4}$$

Solve the equation t=\frac{-10±6}{-4} when ± is plus and when ± is minus.

$$t=1 t=4$$

Since x=t^{2}, the solutions are obtained by evaluating x=±\sqrt{t} for each t.

$$x=1 x=-1 x=2 x=-2$$

Variable x cannot be equal to any of the values 1,-1.

$$x=-2 x=2$$