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

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

Solve for x

$$x=\frac{\sqrt{5}}{5}\approx 0.447213595$$

$$x=-\frac{\sqrt{5}}{5}\approx -0.447213595$$

Steps by Finding Square Root

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

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

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

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

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

$$x^{3}+3x^{2}+3x+1-x^{3}+3x^{2}-3x+1=x^{2}+3$$

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

$$3x^{2}+3x+1+3x^{2}-3x+1=x^{2}+3$$

Combine 3x^{2} and 3x^{2} to get 6x^{2}.

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

Combine 3x and -3x to get 0.

$$6x^{2}+1+1=x^{2}+3$$

Add 1 and 1 to get 2.

$$6x^{2}+2=x^{2}+3$$

Subtract x^{2} from both sides.

$$6x^{2}+2-x^{2}=3$$

Combine 6x^{2} and -x^{2} to get 5x^{2}.

$$5x^{2}+2=3$$

Subtract 2 from both sides.

$$5x^{2}=3-2$$

Subtract 2 from 3 to get 1.

$$5x^{2}=1$$

Divide both sides by 5.

$$x^{2}=\frac{1}{5}$$

Take the square root of both sides of the equation.

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

Steps Using the Quadratic Formula

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

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

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

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

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

$$x^{3}+3x^{2}+3x+1-x^{3}+3x^{2}-3x+1=x^{2}+3$$

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

$$3x^{2}+3x+1+3x^{2}-3x+1=x^{2}+3$$

Combine 3x^{2} and 3x^{2} to get 6x^{2}.

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

Combine 3x and -3x to get 0.

$$6x^{2}+1+1=x^{2}+3$$

Add 1 and 1 to get 2.

$$6x^{2}+2=x^{2}+3$$

Subtract x^{2} from both sides.

$$6x^{2}+2-x^{2}=3$$

Combine 6x^{2} and -x^{2} to get 5x^{2}.

$$5x^{2}+2=3$$

Subtract 3 from both sides.

$$5x^{2}+2-3=0$$

Subtract 3 from 2 to get -1.

$$5x^{2}-1=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 5 for a, 0 for b, and -1 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{0±\sqrt{0^{2}-4\times 5\left(-1\right)}}{2\times 5}$$

Square 0.

$$x=\frac{0±\sqrt{-4\times 5\left(-1\right)}}{2\times 5}$$

Multiply -4 times 5.

$$x=\frac{0±\sqrt{-20\left(-1\right)}}{2\times 5}$$

Multiply -20 times -1.

$$x=\frac{0±\sqrt{20}}{2\times 5}$$

Take the square root of 20.

$$x=\frac{0±2\sqrt{5}}{2\times 5}$$

Multiply 2 times 5.

$$x=\frac{0±2\sqrt{5}}{10}$$

Now solve the equation x=\frac{0±2\sqrt{5}}{10} when ± is plus.

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

Now solve the equation x=\frac{0±2\sqrt{5}}{10} when ± is minus.

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

The equation is now solved.

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