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

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

Evaluate

$$10-2x^{2}$$

Solution Steps

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

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

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

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

Consider \left(x^{2}-3\right)\left(x^{2}+3\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 3.

$$x^{4}-2x^{2}+1-\left(\left(x^{2}\right)^{2}-9\right)$$

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

$$x^{4}-2x^{2}+1-\left(x^{4}-9\right)$$

To find the opposite of x^{4}-9, find the opposite of each term.

$$x^{4}-2x^{2}+1-x^{4}+9$$

Combine x^{4} and -x^{4} to get 0.

$$-2x^{2}+1+9$$

Add 1 and 9 to get 10.

$$-2x^{2}+10$$

Expand

$$10-2x^{2}$$

Solution Steps

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

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

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

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

Consider \left(x^{2}-3\right)\left(x^{2}+3\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 3.

$$x^{4}-2x^{2}+1-\left(\left(x^{2}\right)^{2}-9\right)$$

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

$$x^{4}-2x^{2}+1-\left(x^{4}-9\right)$$

To find the opposite of x^{4}-9, find the opposite of each term.

$$x^{4}-2x^{2}+1-x^{4}+9$$

Combine x^{4} and -x^{4} to get 0.

$$-2x^{2}+1+9$$

Add 1 and 9 to get 10.

$$-2x^{2}+10$$