Solve (1-2a^{2})(1+2a^{2})+(5a^{2}-1)^{2}-2(1-4a^{2})^2-[-2a^4-(3a^2-1)^2]

$$( 1 - 2 a ^ { 2 } ) ( 1 + 2 a ^ { 2 } ) + ( 5 a ^ { 2 } - 1 ) ^ { 2 } - 2 ( 1 - 4 a ^ { 2 } ) ^ { 2 } - [ - 2 a ^ { 4 } - ( 3 a ^ { 2 } - 1 ) ^ { 2 } ]$$

Evaluate

$$1$$

Solution Steps

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

$$1-\left(2a^{2}\right)^{2}+\left(5a^{2}-1\right)^{2}-2\left(1-4a^{2}\right)^{2}-\left(-2a^{4}-\left(3a^{2}-1\right)^{2}\right)$$

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

$$1-2^{2}\left(a^{2}\right)^{2}+\left(5a^{2}-1\right)^{2}-2\left(1-4a^{2}\right)^{2}-\left(-2a^{4}-\left(3a^{2}-1\right)^{2}\right)$$

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

$$1-2^{2}a^{4}+\left(5a^{2}-1\right)^{2}-2\left(1-4a^{2}\right)^{2}-\left(-2a^{4}-\left(3a^{2}-1\right)^{2}\right)$$

Calculate 2 to the power of 2 and get 4.

$$1-4a^{4}+\left(5a^{2}-1\right)^{2}-2\left(1-4a^{2}\right)^{2}-\left(-2a^{4}-\left(3a^{2}-1\right)^{2}\right)$$

Use binomial theorem \left(p-q\right)^{2}=p^{2}-2pq+q^{2} to expand \left(5a^{2}-1\right)^{2}.

$$1-4a^{4}+25\left(a^{2}\right)^{2}-10a^{2}+1-2\left(1-4a^{2}\right)^{2}-\left(-2a^{4}-\left(3a^{2}-1\right)^{2}\right)$$

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

$$1-4a^{4}+25a^{4}-10a^{2}+1-2\left(1-4a^{2}\right)^{2}-\left(-2a^{4}-\left(3a^{2}-1\right)^{2}\right)$$

Combine -4a^{4} and 25a^{4} to get 21a^{4}.

$$1+21a^{4}-10a^{2}+1-2\left(1-4a^{2}\right)^{2}-\left(-2a^{4}-\left(3a^{2}-1\right)^{2}\right)$$

Add 1 and 1 to get 2.

$$2+21a^{4}-10a^{2}-2\left(1-4a^{2}\right)^{2}-\left(-2a^{4}-\left(3a^{2}-1\right)^{2}\right)$$

Use binomial theorem \left(p-q\right)^{2}=p^{2}-2pq+q^{2} to expand \left(1-4a^{2}\right)^{2}.

$$2+21a^{4}-10a^{2}-2\left(1-8a^{2}+16\left(a^{2}\right)^{2}\right)-\left(-2a^{4}-\left(3a^{2}-1\right)^{2}\right)$$

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

$$2+21a^{4}-10a^{2}-2\left(1-8a^{2}+16a^{4}\right)-\left(-2a^{4}-\left(3a^{2}-1\right)^{2}\right)$$

Use the distributive property to multiply -2 by 1-8a^{2}+16a^{4}.

$$2+21a^{4}-10a^{2}-2+16a^{2}-32a^{4}-\left(-2a^{4}-\left(3a^{2}-1\right)^{2}\right)$$

Subtract 2 from 2 to get 0.

$$21a^{4}-10a^{2}+16a^{2}-32a^{4}-\left(-2a^{4}-\left(3a^{2}-1\right)^{2}\right)$$

Combine -10a^{2} and 16a^{2} to get 6a^{2}.

$$21a^{4}+6a^{2}-32a^{4}-\left(-2a^{4}-\left(3a^{2}-1\right)^{2}\right)$$

Combine 21a^{4} and -32a^{4} to get -11a^{4}.

$$-11a^{4}+6a^{2}-\left(-2a^{4}-\left(3a^{2}-1\right)^{2}\right)$$

Use binomial theorem \left(p-q\right)^{2}=p^{2}-2pq+q^{2} to expand \left(3a^{2}-1\right)^{2}.

$$-11a^{4}+6a^{2}-\left(-2a^{4}-\left(9\left(a^{2}\right)^{2}-6a^{2}+1\right)\right)$$

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

$$-11a^{4}+6a^{2}-\left(-2a^{4}-\left(9a^{4}-6a^{2}+1\right)\right)$$

To find the opposite of 9a^{4}-6a^{2}+1, find the opposite of each term.

$$-11a^{4}+6a^{2}-\left(-2a^{4}-9a^{4}+6a^{2}-1\right)$$

Combine -2a^{4} and -9a^{4} to get -11a^{4}.

$$-11a^{4}+6a^{2}-\left(-11a^{4}+6a^{2}-1\right)$$

To find the opposite of -11a^{4}+6a^{2}-1, find the opposite of each term.

$$-11a^{4}+6a^{2}+11a^{4}-6a^{2}+1$$

Combine -11a^{4} and 11a^{4} to get 0.

$$6a^{2}-6a^{2}+1$$

Combine 6a^{2} and -6a^{2} to get 0.

$$1$$

Factor

$$1$$