Solve (a^{2}-1)(a^{2}-2)-[(3a^2-2)^2+(5+2a^2)(5-2a^2)]=

$$( a ^ { 2 } - 1 ) ( a ^ { 2 } - 2 ) - [ ( 3 a ^ { 2 } - 2 ) ^ { 2 } + ( 5 + 2 a ^ { 2 } ) ( 5 - 2 a ^ { 2 } ) ] =$$

Evaluate

$$-4a^{4}+9a^{2}-27$$

Solution Steps

Use the distributive property to multiply a^{2}-1 by a^{2}-2 and combine like terms.

$$a^{4}-3a^{2}+2-\left(\left(3a^{2}-2\right)^{2}+\left(5+2a^{2}\right)\left(5-2a^{2}\right)\right)$$

Use binomial theorem \left(p-q\right)^{2}=p^{2}-2pq+q^{2} to expand \left(3a^{2}-2\right)^{2}.

$$a^{4}-3a^{2}+2-\left(9\left(a^{2}\right)^{2}-12a^{2}+4+\left(5+2a^{2}\right)\left(5-2a^{2}\right)\right)$$

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

$$a^{4}-3a^{2}+2-\left(9a^{4}-12a^{2}+4+\left(5+2a^{2}\right)\left(5-2a^{2}\right)\right)$$

Consider \left(5+2a^{2}\right)\left(5-2a^{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 5.

$$a^{4}-3a^{2}+2-\left(9a^{4}-12a^{2}+4+25-\left(2a^{2}\right)^{2}\right)$$

Expand \left(2a^{2}\right)^{2}.

$$a^{4}-3a^{2}+2-\left(9a^{4}-12a^{2}+4+25-2^{2}\left(a^{2}\right)^{2}\right)$$

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

$$a^{4}-3a^{2}+2-\left(9a^{4}-12a^{2}+4+25-2^{2}a^{4}\right)$$

Calculate 2 to the power of 2 and get 4.

$$a^{4}-3a^{2}+2-\left(9a^{4}-12a^{2}+4+25-4a^{4}\right)$$

Add 4 and 25 to get 29.

$$a^{4}-3a^{2}+2-\left(9a^{4}-12a^{2}+29-4a^{4}\right)$$

Combine 9a^{4} and -4a^{4} to get 5a^{4}.

$$a^{4}-3a^{2}+2-\left(5a^{4}-12a^{2}+29\right)$$

To find the opposite of 5a^{4}-12a^{2}+29, find the opposite of each term.

$$a^{4}-3a^{2}+2-5a^{4}+12a^{2}-29$$

Combine a^{4} and -5a^{4} to get -4a^{4}.

$$-4a^{4}-3a^{2}+2+12a^{2}-29$$

Combine -3a^{2} and 12a^{2} to get 9a^{2}.

$$-4a^{4}+9a^{2}+2-29$$

Subtract 29 from 2 to get -27.

$$-4a^{4}+9a^{2}-27$$

Expand

$$-4a^{4}+9a^{2}-27$$

Solution Steps

Use the distributive property to multiply a^{2}-1 by a^{2}-2 and combine like terms.

$$a^{4}-3a^{2}+2-\left(\left(3a^{2}-2\right)^{2}+\left(5+2a^{2}\right)\left(5-2a^{2}\right)\right)$$

Use binomial theorem \left(p-q\right)^{2}=p^{2}-2pq+q^{2} to expand \left(3a^{2}-2\right)^{2}.

$$a^{4}-3a^{2}+2-\left(9\left(a^{2}\right)^{2}-12a^{2}+4+\left(5+2a^{2}\right)\left(5-2a^{2}\right)\right)$$

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

$$a^{4}-3a^{2}+2-\left(9a^{4}-12a^{2}+4+\left(5+2a^{2}\right)\left(5-2a^{2}\right)\right)$$

Consider \left(5+2a^{2}\right)\left(5-2a^{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 5.

$$a^{4}-3a^{2}+2-\left(9a^{4}-12a^{2}+4+25-\left(2a^{2}\right)^{2}\right)$$

Expand \left(2a^{2}\right)^{2}.

$$a^{4}-3a^{2}+2-\left(9a^{4}-12a^{2}+4+25-2^{2}\left(a^{2}\right)^{2}\right)$$

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

$$a^{4}-3a^{2}+2-\left(9a^{4}-12a^{2}+4+25-2^{2}a^{4}\right)$$

Calculate 2 to the power of 2 and get 4.

$$a^{4}-3a^{2}+2-\left(9a^{4}-12a^{2}+4+25-4a^{4}\right)$$

Add 4 and 25 to get 29.

$$a^{4}-3a^{2}+2-\left(9a^{4}-12a^{2}+29-4a^{4}\right)$$

Combine 9a^{4} and -4a^{4} to get 5a^{4}.

$$a^{4}-3a^{2}+2-\left(5a^{4}-12a^{2}+29\right)$$

To find the opposite of 5a^{4}-12a^{2}+29, find the opposite of each term.

$$a^{4}-3a^{2}+2-5a^{4}+12a^{2}-29$$

Combine a^{4} and -5a^{4} to get -4a^{4}.

$$-4a^{4}-3a^{2}+2+12a^{2}-29$$

Combine -3a^{2} and 12a^{2} to get 9a^{2}.

$$-4a^{4}+9a^{2}+2-29$$

Subtract 29 from 2 to get -27.

$$-4a^{4}+9a^{2}-27$$