Solve [(-2a+1/2b)(-2a-1/2b)]^{2}(4a^{2}+1/4b^{2})^2-(16a^4+1/16b^4)^2=

$$[ ( - 2 a + \frac { 1 } { 2 } b ) ( - 2 a - \frac { 1 } { 2 } b ) ] ^ { 2 } ( 4 a ^ { 2 } + \frac { 1 } { 4 } b ^ { 2 } ) ^ { 2 } - ( 16 a ^ { 4 } + \frac { 1 } { 16 } b ^ { 4 } ) ^ { 2 } =$$

Evaluate

$$-4\left(ab\right)^{4}$$

Solution Steps

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

$$\left(\left(-2a\right)^{2}-\left(\frac{1}{2}b\right)^{2}\right)^{2}\left(4a^{2}+\frac{1}{4}b^{2}\right)^{2}-\left(16a^{4}+\frac{1}{16}b^{4}\right)^{2}$$

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

$$\left(\left(-2\right)^{2}a^{2}-\left(\frac{1}{2}b\right)^{2}\right)^{2}\left(4a^{2}+\frac{1}{4}b^{2}\right)^{2}-\left(16a^{4}+\frac{1}{16}b^{4}\right)^{2}$$

Calculate -2 to the power of 2 and get 4.

$$\left(4a^{2}-\left(\frac{1}{2}b\right)^{2}\right)^{2}\left(4a^{2}+\frac{1}{4}b^{2}\right)^{2}-\left(16a^{4}+\frac{1}{16}b^{4}\right)^{2}$$

Expand \left(\frac{1}{2}b\right)^{2}.

$$\left(4a^{2}-\left(\frac{1}{2}\right)^{2}b^{2}\right)^{2}\left(4a^{2}+\frac{1}{4}b^{2}\right)^{2}-\left(16a^{4}+\frac{1}{16}b^{4}\right)^{2}$$

Calculate \frac{1}{2} to the power of 2 and get \frac{1}{4}.

$$\left(4a^{2}-\frac{1}{4}b^{2}\right)^{2}\left(4a^{2}+\frac{1}{4}b^{2}\right)^{2}-\left(16a^{4}+\frac{1}{16}b^{4}\right)^{2}$$

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

$$\left(16\left(a^{2}\right)^{2}-2a^{2}b^{2}+\frac{1}{16}\left(b^{2}\right)^{2}\right)\left(4a^{2}+\frac{1}{4}b^{2}\right)^{2}-\left(16a^{4}+\frac{1}{16}b^{4}\right)^{2}$$

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

$$\left(16a^{4}-2a^{2}b^{2}+\frac{1}{16}\left(b^{2}\right)^{2}\right)\left(4a^{2}+\frac{1}{4}b^{2}\right)^{2}-\left(16a^{4}+\frac{1}{16}b^{4}\right)^{2}$$

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

$$\left(16a^{4}-2a^{2}b^{2}+\frac{1}{16}b^{4}\right)\left(4a^{2}+\frac{1}{4}b^{2}\right)^{2}-\left(16a^{4}+\frac{1}{16}b^{4}\right)^{2}$$

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

$$\left(16a^{4}-2a^{2}b^{2}+\frac{1}{16}b^{4}\right)\left(16\left(a^{2}\right)^{2}+2a^{2}b^{2}+\frac{1}{16}\left(b^{2}\right)^{2}\right)-\left(16a^{4}+\frac{1}{16}b^{4}\right)^{2}$$

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

$$\left(16a^{4}-2a^{2}b^{2}+\frac{1}{16}b^{4}\right)\left(16a^{4}+2a^{2}b^{2}+\frac{1}{16}\left(b^{2}\right)^{2}\right)-\left(16a^{4}+\frac{1}{16}b^{4}\right)^{2}$$

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

$$\left(16a^{4}-2a^{2}b^{2}+\frac{1}{16}b^{4}\right)\left(16a^{4}+2a^{2}b^{2}+\frac{1}{16}b^{4}\right)-\left(16a^{4}+\frac{1}{16}b^{4}\right)^{2}$$

Use the distributive property to multiply 16a^{4}-2a^{2}b^{2}+\frac{1}{16}b^{4} by 16a^{4}+2a^{2}b^{2}+\frac{1}{16}b^{4} and combine like terms.

$$256a^{8}-2a^{4}b^{4}+\frac{1}{256}b^{8}-\left(16a^{4}+\frac{1}{16}b^{4}\right)^{2}$$

Use binomial theorem \left(p+q\right)^{2}=p^{2}+2pq+q^{2} to expand \left(16a^{4}+\frac{1}{16}b^{4}\right)^{2}.

$$256a^{8}-2a^{4}b^{4}+\frac{1}{256}b^{8}-\left(256\left(a^{4}\right)^{2}+2a^{4}b^{4}+\frac{1}{256}\left(b^{4}\right)^{2}\right)$$

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

$$256a^{8}-2a^{4}b^{4}+\frac{1}{256}b^{8}-\left(256a^{8}+2a^{4}b^{4}+\frac{1}{256}\left(b^{4}\right)^{2}\right)$$

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

$$256a^{8}-2a^{4}b^{4}+\frac{1}{256}b^{8}-\left(256a^{8}+2a^{4}b^{4}+\frac{1}{256}b^{8}\right)$$

To find the opposite of 256a^{8}+2a^{4}b^{4}+\frac{1}{256}b^{8}, find the opposite of each term.

$$256a^{8}-2a^{4}b^{4}+\frac{1}{256}b^{8}-256a^{8}-2a^{4}b^{4}-\frac{1}{256}b^{8}$$

Combine 256a^{8} and -256a^{8} to get 0.

$$-2a^{4}b^{4}+\frac{1}{256}b^{8}-2a^{4}b^{4}-\frac{1}{256}b^{8}$$

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

$$-4a^{4}b^{4}+\frac{1}{256}b^{8}-\frac{1}{256}b^{8}$$

Combine \frac{1}{256}b^{8} and -\frac{1}{256}b^{8} to get 0.

$$-4a^{4}b^{4}$$

Expand

$$-4\left(ab\right)^{4}$$

Solution Steps

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

$$\left(\left(-2a\right)^{2}-\left(\frac{1}{2}b\right)^{2}\right)^{2}\left(4a^{2}+\frac{1}{4}b^{2}\right)^{2}-\left(16a^{4}+\frac{1}{16}b^{4}\right)^{2}$$

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

$$\left(\left(-2\right)^{2}a^{2}-\left(\frac{1}{2}b\right)^{2}\right)^{2}\left(4a^{2}+\frac{1}{4}b^{2}\right)^{2}-\left(16a^{4}+\frac{1}{16}b^{4}\right)^{2}$$

Calculate -2 to the power of 2 and get 4.

$$\left(4a^{2}-\left(\frac{1}{2}b\right)^{2}\right)^{2}\left(4a^{2}+\frac{1}{4}b^{2}\right)^{2}-\left(16a^{4}+\frac{1}{16}b^{4}\right)^{2}$$

Expand \left(\frac{1}{2}b\right)^{2}.

$$\left(4a^{2}-\left(\frac{1}{2}\right)^{2}b^{2}\right)^{2}\left(4a^{2}+\frac{1}{4}b^{2}\right)^{2}-\left(16a^{4}+\frac{1}{16}b^{4}\right)^{2}$$

Calculate \frac{1}{2} to the power of 2 and get \frac{1}{4}.

$$\left(4a^{2}-\frac{1}{4}b^{2}\right)^{2}\left(4a^{2}+\frac{1}{4}b^{2}\right)^{2}-\left(16a^{4}+\frac{1}{16}b^{4}\right)^{2}$$

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

$$\left(16\left(a^{2}\right)^{2}-2a^{2}b^{2}+\frac{1}{16}\left(b^{2}\right)^{2}\right)\left(4a^{2}+\frac{1}{4}b^{2}\right)^{2}-\left(16a^{4}+\frac{1}{16}b^{4}\right)^{2}$$

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

$$\left(16a^{4}-2a^{2}b^{2}+\frac{1}{16}\left(b^{2}\right)^{2}\right)\left(4a^{2}+\frac{1}{4}b^{2}\right)^{2}-\left(16a^{4}+\frac{1}{16}b^{4}\right)^{2}$$

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

$$\left(16a^{4}-2a^{2}b^{2}+\frac{1}{16}b^{4}\right)\left(4a^{2}+\frac{1}{4}b^{2}\right)^{2}-\left(16a^{4}+\frac{1}{16}b^{4}\right)^{2}$$

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

$$\left(16a^{4}-2a^{2}b^{2}+\frac{1}{16}b^{4}\right)\left(16\left(a^{2}\right)^{2}+2a^{2}b^{2}+\frac{1}{16}\left(b^{2}\right)^{2}\right)-\left(16a^{4}+\frac{1}{16}b^{4}\right)^{2}$$

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

$$\left(16a^{4}-2a^{2}b^{2}+\frac{1}{16}b^{4}\right)\left(16a^{4}+2a^{2}b^{2}+\frac{1}{16}\left(b^{2}\right)^{2}\right)-\left(16a^{4}+\frac{1}{16}b^{4}\right)^{2}$$

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

$$\left(16a^{4}-2a^{2}b^{2}+\frac{1}{16}b^{4}\right)\left(16a^{4}+2a^{2}b^{2}+\frac{1}{16}b^{4}\right)-\left(16a^{4}+\frac{1}{16}b^{4}\right)^{2}$$

Use the distributive property to multiply 16a^{4}-2a^{2}b^{2}+\frac{1}{16}b^{4} by 16a^{4}+2a^{2}b^{2}+\frac{1}{16}b^{4} and combine like terms.

$$256a^{8}-2a^{4}b^{4}+\frac{1}{256}b^{8}-\left(16a^{4}+\frac{1}{16}b^{4}\right)^{2}$$

Use binomial theorem \left(p+q\right)^{2}=p^{2}+2pq+q^{2} to expand \left(16a^{4}+\frac{1}{16}b^{4}\right)^{2}.

$$256a^{8}-2a^{4}b^{4}+\frac{1}{256}b^{8}-\left(256\left(a^{4}\right)^{2}+2a^{4}b^{4}+\frac{1}{256}\left(b^{4}\right)^{2}\right)$$

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

$$256a^{8}-2a^{4}b^{4}+\frac{1}{256}b^{8}-\left(256a^{8}+2a^{4}b^{4}+\frac{1}{256}\left(b^{4}\right)^{2}\right)$$

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

$$256a^{8}-2a^{4}b^{4}+\frac{1}{256}b^{8}-\left(256a^{8}+2a^{4}b^{4}+\frac{1}{256}b^{8}\right)$$

To find the opposite of 256a^{8}+2a^{4}b^{4}+\frac{1}{256}b^{8}, find the opposite of each term.

$$256a^{8}-2a^{4}b^{4}+\frac{1}{256}b^{8}-256a^{8}-2a^{4}b^{4}-\frac{1}{256}b^{8}$$

Combine 256a^{8} and -256a^{8} to get 0.

$$-2a^{4}b^{4}+\frac{1}{256}b^{8}-2a^{4}b^{4}-\frac{1}{256}b^{8}$$

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

$$-4a^{4}b^{4}+\frac{1}{256}b^{8}-\frac{1}{256}b^{8}$$

Combine \frac{1}{256}b^{8} and -\frac{1}{256}b^{8} to get 0.

$$-4a^{4}b^{4}$$