Solve ((x^{2}+1)(x^{2}-1))^{2}-(x^{2}-2)^2+[(x+1)(x-1)]^2-(x^2+2)(x^2-2)

$$( ( x ^ { 2 } + 1 ) ( x ^ { 2 } - 1 ) ) ^ { 2 } - ( x ^ { 2 } - 2 ) ^ { 2 } + [ ( x + 1 ) ( x - 1 ) ] ^ { 2 } - ( x ^ { 2 } + 2 ) ( x ^ { 2 } - 2 )$$

Evaluate

$$x^{8}-3x^{4}+2x^{2}+2$$

Solution Steps

Consider \left(x^{2}+1\right)\left(x^{2}-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.

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

To raise a power to another power, multiply the exponents. Multiply 2 and 2 to get 4.

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

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x^{4}-1\right)^{2}.

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

To raise a power to another power, multiply the exponents. Multiply 4 and 2 to get 8.

$$x^{8}-2x^{4}+1-\left(x^{2}-2\right)^{2}+\left(\left(x+1\right)\left(x-1\right)\right)^{2}-\left(x^{2}+2\right)\left(x^{2}-2\right)$$

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

$$x^{8}-2x^{4}+1-\left(\left(x^{2}\right)^{2}-4x^{2}+4\right)+\left(\left(x+1\right)\left(x-1\right)\right)^{2}-\left(x^{2}+2\right)\left(x^{2}-2\right)$$

To raise a power to another power, multiply the exponents. Multiply 2 and 2 to get 4.

$$x^{8}-2x^{4}+1-\left(x^{4}-4x^{2}+4\right)+\left(\left(x+1\right)\left(x-1\right)\right)^{2}-\left(x^{2}+2\right)\left(x^{2}-2\right)$$

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

$$x^{8}-2x^{4}+1-x^{4}+4x^{2}-4+\left(\left(x+1\right)\left(x-1\right)\right)^{2}-\left(x^{2}+2\right)\left(x^{2}-2\right)$$

Combine -2x^{4} and -x^{4} to get -3x^{4}.

$$x^{8}-3x^{4}+1+4x^{2}-4+\left(\left(x+1\right)\left(x-1\right)\right)^{2}-\left(x^{2}+2\right)\left(x^{2}-2\right)$$

Subtract 4 from 1 to get -3.

$$x^{8}-3x^{4}-3+4x^{2}+\left(\left(x+1\right)\left(x-1\right)\right)^{2}-\left(x^{2}+2\right)\left(x^{2}-2\right)$$

Consider \left(x+1\right)\left(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}. Square 1.

$$x^{8}-3x^{4}-3+4x^{2}+\left(x^{2}-1\right)^{2}-\left(x^{2}+2\right)\left(x^{2}-2\right)$$

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

$$x^{8}-3x^{4}-3+4x^{2}+\left(x^{2}\right)^{2}-2x^{2}+1-\left(x^{2}+2\right)\left(x^{2}-2\right)$$

To raise a power to another power, multiply the exponents. Multiply 2 and 2 to get 4.

$$x^{8}-3x^{4}-3+4x^{2}+x^{4}-2x^{2}+1-\left(x^{2}+2\right)\left(x^{2}-2\right)$$

Combine -3x^{4} and x^{4} to get -2x^{4}.

$$x^{8}-2x^{4}-3+4x^{2}-2x^{2}+1-\left(x^{2}+2\right)\left(x^{2}-2\right)$$

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

$$x^{8}-2x^{4}-3+2x^{2}+1-\left(x^{2}+2\right)\left(x^{2}-2\right)$$

Add -3 and 1 to get -2.

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

Consider \left(x^{2}+2\right)\left(x^{2}-2\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 2.

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

To raise a power to another power, multiply the exponents. Multiply 2 and 2 to get 4.

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

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

$$x^{8}-2x^{4}-2+2x^{2}-x^{4}+4$$

Combine -2x^{4} and -x^{4} to get -3x^{4}.

$$x^{8}-3x^{4}-2+2x^{2}+4$$

Add -2 and 4 to get 2.

$$x^{8}-3x^{4}+2+2x^{2}$$

Expand

$$x^{8}-3x^{4}+2x^{2}+2$$

Solution Steps

Consider \left(x^{2}+1\right)\left(x^{2}-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.

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

To raise a power to another power, multiply the exponents. Multiply 2 and 2 to get 4.

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

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x^{4}-1\right)^{2}.

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

To raise a power to another power, multiply the exponents. Multiply 4 and 2 to get 8.

$$x^{8}-2x^{4}+1-\left(x^{2}-2\right)^{2}+\left(\left(x+1\right)\left(x-1\right)\right)^{2}-\left(x^{2}+2\right)\left(x^{2}-2\right)$$

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

$$x^{8}-2x^{4}+1-\left(\left(x^{2}\right)^{2}-4x^{2}+4\right)+\left(\left(x+1\right)\left(x-1\right)\right)^{2}-\left(x^{2}+2\right)\left(x^{2}-2\right)$$

To raise a power to another power, multiply the exponents. Multiply 2 and 2 to get 4.

$$x^{8}-2x^{4}+1-\left(x^{4}-4x^{2}+4\right)+\left(\left(x+1\right)\left(x-1\right)\right)^{2}-\left(x^{2}+2\right)\left(x^{2}-2\right)$$

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

$$x^{8}-2x^{4}+1-x^{4}+4x^{2}-4+\left(\left(x+1\right)\left(x-1\right)\right)^{2}-\left(x^{2}+2\right)\left(x^{2}-2\right)$$

Combine -2x^{4} and -x^{4} to get -3x^{4}.

$$x^{8}-3x^{4}+1+4x^{2}-4+\left(\left(x+1\right)\left(x-1\right)\right)^{2}-\left(x^{2}+2\right)\left(x^{2}-2\right)$$

Subtract 4 from 1 to get -3.

$$x^{8}-3x^{4}-3+4x^{2}+\left(\left(x+1\right)\left(x-1\right)\right)^{2}-\left(x^{2}+2\right)\left(x^{2}-2\right)$$

Consider \left(x+1\right)\left(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}. Square 1.

$$x^{8}-3x^{4}-3+4x^{2}+\left(x^{2}-1\right)^{2}-\left(x^{2}+2\right)\left(x^{2}-2\right)$$

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

$$x^{8}-3x^{4}-3+4x^{2}+\left(x^{2}\right)^{2}-2x^{2}+1-\left(x^{2}+2\right)\left(x^{2}-2\right)$$

To raise a power to another power, multiply the exponents. Multiply 2 and 2 to get 4.

$$x^{8}-3x^{4}-3+4x^{2}+x^{4}-2x^{2}+1-\left(x^{2}+2\right)\left(x^{2}-2\right)$$

Combine -3x^{4} and x^{4} to get -2x^{4}.

$$x^{8}-2x^{4}-3+4x^{2}-2x^{2}+1-\left(x^{2}+2\right)\left(x^{2}-2\right)$$

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

$$x^{8}-2x^{4}-3+2x^{2}+1-\left(x^{2}+2\right)\left(x^{2}-2\right)$$

Add -3 and 1 to get -2.

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

Consider \left(x^{2}+2\right)\left(x^{2}-2\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 2.

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

To raise a power to another power, multiply the exponents. Multiply 2 and 2 to get 4.

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

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

$$x^{8}-2x^{4}-2+2x^{2}-x^{4}+4$$

Combine -2x^{4} and -x^{4} to get -3x^{4}.

$$x^{8}-3x^{4}-2+2x^{2}+4$$

Add -2 and 4 to get 2.

$$x^{8}-3x^{4}+2+2x^{2}$$