Solve (x^n+x-1)(x^n+x+1)-(x^n-1)(x^n+1)=

$$( x ^ { n } + x - 1 ) ( x ^ { n } + x + 1 ) - ( x ^ { n } - 1 ) ( x ^ { n } + 1 ) =$$

Differentiate w.r.t. x

$$2\left(\left(n+1\right)x^{n}+x\right)$$

Steps Using Derivative Rule for Sum

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

$$\frac{\mathrm{d}}{\mathrm{d}x}(\left(x^{n}+x\right)^{2}-1-\left(x^{n}-1\right)\left(x^{n}+1\right))$$

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

$$\frac{\mathrm{d}}{\mathrm{d}x}(\left(x^{n}\right)^{2}+2x^{n}x+x^{2}-1-\left(x^{n}-1\right)\left(x^{n}+1\right))$$

Consider \left(x^{n}-1\right)\left(x^{n}+1\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.

$$\frac{\mathrm{d}}{\mathrm{d}x}(\left(x^{n}\right)^{2}+2x^{n}x+x^{2}-1-\left(\left(x^{n}\right)^{2}-1\right))$$

To find the opposite of \left(x^{n}\right)^{2}-1, find the opposite of each term.

$$\frac{\mathrm{d}}{\mathrm{d}x}(\left(x^{n}\right)^{2}+2x^{n}x+x^{2}-1-\left(x^{n}\right)^{2}+1)$$

Combine \left(x^{n}\right)^{2} and -\left(x^{n}\right)^{2} to get 0.

$$\frac{\mathrm{d}}{\mathrm{d}x}(2x^{n}x+x^{2}-1+1)$$

Add -1 and 1 to get 0.

$$\frac{\mathrm{d}}{\mathrm{d}x}(2x^{n}x+x^{2})$$

The derivative of a polynomial is the sum of the derivatives of its terms. The derivative of a constant term is 0. The derivative of ax^{n} is nax^{n-1}.

$$2x^{n}x^{1-1}+2x^{2-1}$$

Subtract 1 from 1.

$$2x^{n}x^{0}+2x^{2-1}$$

Subtract 1 from 2.

$$2x^{n}x^{0}+2x^{1}$$

For any term t, t^{1}=t.

$$2x^{n}x^{0}+2x$$

For any term t except 0, t^{0}=1.

$$2x^{n}\times 1+2x$$

For any term t, t\times 1=t and 1t=t.

$$2x^{n}+2x$$

Evaluate

$$x\left(x+2x^{n}\right)$$