Solve (-1/8a^{3}-a+4)(1/8a^{3}-a-4)+(1/4a^2-a)^3-(3a-8)(3a+8)-3(1/2a^2-4)^2

$$( - \frac { 1 } { 8 } a ^ { 3 } - a + 4 ) ( \frac { 1 } { 8 } a ^ { 3 } - a - 4 ) + ( \frac { 1 } { 4 } a ^ { 2 } - a ) ^ { 3 } - ( 3 a - 8 ) ( 3 a + 8 ) - 3 ( \frac { 1 } { 2 } a ^ { 2 } - 4 ) ^ { 2 }$$

Evaluate

$$-\frac{3a^{5}}{16}+4a^{2}$$

Solution Steps

Use the distributive property to multiply -\frac{1}{8}a^{3}-a+4 by \frac{1}{8}a^{3}-a-4 and combine like terms.

$$-\frac{1}{64}a^{6}+a^{3}+a^{2}-16+\left(\frac{1}{4}a^{2}-a\right)^{3}-\left(3a-8\right)\left(3a+8\right)-3\left(\frac{1}{2}a^{2}-4\right)^{2}$$

Use binomial theorem \left(p-q\right)^{3}=p^{3}-3p^{2}q+3pq^{2}-q^{3} to expand \left(\frac{1}{4}a^{2}-a\right)^{3}.

$$-\frac{1}{64}a^{6}+a^{3}+a^{2}-16+\frac{1}{64}\left(a^{2}\right)^{3}-\frac{3}{16}\left(a^{2}\right)^{2}a+\frac{3}{4}a^{2}a^{2}-a^{3}-\left(3a-8\right)\left(3a+8\right)-3\left(\frac{1}{2}a^{2}-4\right)^{2}$$

To raise a power to another power, multiply the exponents. Multiply 2 and 3 to get 6.

$$-\frac{1}{64}a^{6}+a^{3}+a^{2}-16+\frac{1}{64}a^{6}-\frac{3}{16}\left(a^{2}\right)^{2}a+\frac{3}{4}a^{2}a^{2}-a^{3}-\left(3a-8\right)\left(3a+8\right)-3\left(\frac{1}{2}a^{2}-4\right)^{2}$$

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

$$-\frac{1}{64}a^{6}+a^{3}+a^{2}-16+\frac{1}{64}a^{6}-\frac{3}{16}a^{4}a+\frac{3}{4}a^{2}a^{2}-a^{3}-\left(3a-8\right)\left(3a+8\right)-3\left(\frac{1}{2}a^{2}-4\right)^{2}$$

To multiply powers of the same base, add their exponents. Add 4 and 1 to get 5.

$$-\frac{1}{64}a^{6}+a^{3}+a^{2}-16+\frac{1}{64}a^{6}-\frac{3}{16}a^{5}+\frac{3}{4}a^{2}a^{2}-a^{3}-\left(3a-8\right)\left(3a+8\right)-3\left(\frac{1}{2}a^{2}-4\right)^{2}$$

To multiply powers of the same base, add their exponents. Add 2 and 2 to get 4.

$$-\frac{1}{64}a^{6}+a^{3}+a^{2}-16+\frac{1}{64}a^{6}-\frac{3}{16}a^{5}+\frac{3}{4}a^{4}-a^{3}-\left(3a-8\right)\left(3a+8\right)-3\left(\frac{1}{2}a^{2}-4\right)^{2}$$

Combine -\frac{1}{64}a^{6} and \frac{1}{64}a^{6} to get 0.

$$a^{3}+a^{2}-16-\frac{3}{16}a^{5}+\frac{3}{4}a^{4}-a^{3}-\left(3a-8\right)\left(3a+8\right)-3\left(\frac{1}{2}a^{2}-4\right)^{2}$$

Combine a^{3} and -a^{3} to get 0.

$$a^{2}-16-\frac{3}{16}a^{5}+\frac{3}{4}a^{4}-\left(3a-8\right)\left(3a+8\right)-3\left(\frac{1}{2}a^{2}-4\right)^{2}$$

Consider \left(3a-8\right)\left(3a+8\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 8.

$$a^{2}-16-\frac{3}{16}a^{5}+\frac{3}{4}a^{4}-\left(\left(3a\right)^{2}-64\right)-3\left(\frac{1}{2}a^{2}-4\right)^{2}$$

Expand \left(3a\right)^{2}.

$$a^{2}-16-\frac{3}{16}a^{5}+\frac{3}{4}a^{4}-\left(3^{2}a^{2}-64\right)-3\left(\frac{1}{2}a^{2}-4\right)^{2}$$

Calculate 3 to the power of 2 and get 9.

$$a^{2}-16-\frac{3}{16}a^{5}+\frac{3}{4}a^{4}-\left(9a^{2}-64\right)-3\left(\frac{1}{2}a^{2}-4\right)^{2}$$

To find the opposite of 9a^{2}-64, find the opposite of each term.

$$a^{2}-16-\frac{3}{16}a^{5}+\frac{3}{4}a^{4}-9a^{2}+64-3\left(\frac{1}{2}a^{2}-4\right)^{2}$$

Combine a^{2} and -9a^{2} to get -8a^{2}.

$$-8a^{2}-16-\frac{3}{16}a^{5}+\frac{3}{4}a^{4}+64-3\left(\frac{1}{2}a^{2}-4\right)^{2}$$

Add -16 and 64 to get 48.

$$-8a^{2}+48-\frac{3}{16}a^{5}+\frac{3}{4}a^{4}-3\left(\frac{1}{2}a^{2}-4\right)^{2}$$

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

$$-8a^{2}+48-\frac{3}{16}a^{5}+\frac{3}{4}a^{4}-3\left(\frac{1}{4}\left(a^{2}\right)^{2}-4a^{2}+16\right)$$

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

$$-8a^{2}+48-\frac{3}{16}a^{5}+\frac{3}{4}a^{4}-3\left(\frac{1}{4}a^{4}-4a^{2}+16\right)$$

Use the distributive property to multiply -3 by \frac{1}{4}a^{4}-4a^{2}+16.

$$-8a^{2}+48-\frac{3}{16}a^{5}+\frac{3}{4}a^{4}-\frac{3}{4}a^{4}+12a^{2}-48$$

Combine \frac{3}{4}a^{4} and -\frac{3}{4}a^{4} to get 0.

$$-8a^{2}+48-\frac{3}{16}a^{5}+12a^{2}-48$$

Combine -8a^{2} and 12a^{2} to get 4a^{2}.

$$4a^{2}+48-\frac{3}{16}a^{5}-48$$

Subtract 48 from 48 to get 0.

$$4a^{2}-\frac{3}{16}a^{5}$$

Expand

$$-\frac{3a^{5}}{16}+4a^{2}$$

Solution Steps

Use the distributive property to multiply -\frac{1}{8}a^{3}-a+4 by \frac{1}{8}a^{3}-a-4 and combine like terms.

$$-\frac{1}{64}a^{6}+a^{3}+a^{2}-16+\left(\frac{1}{4}a^{2}-a\right)^{3}-\left(3a-8\right)\left(3a+8\right)-3\left(\frac{1}{2}a^{2}-4\right)^{2}$$

Use binomial theorem \left(p-q\right)^{3}=p^{3}-3p^{2}q+3pq^{2}-q^{3} to expand \left(\frac{1}{4}a^{2}-a\right)^{3}.

$$-\frac{1}{64}a^{6}+a^{3}+a^{2}-16+\frac{1}{64}\left(a^{2}\right)^{3}-\frac{3}{16}\left(a^{2}\right)^{2}a+\frac{3}{4}a^{2}a^{2}-a^{3}-\left(3a-8\right)\left(3a+8\right)-3\left(\frac{1}{2}a^{2}-4\right)^{2}$$

To raise a power to another power, multiply the exponents. Multiply 2 and 3 to get 6.

$$-\frac{1}{64}a^{6}+a^{3}+a^{2}-16+\frac{1}{64}a^{6}-\frac{3}{16}\left(a^{2}\right)^{2}a+\frac{3}{4}a^{2}a^{2}-a^{3}-\left(3a-8\right)\left(3a+8\right)-3\left(\frac{1}{2}a^{2}-4\right)^{2}$$

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

$$-\frac{1}{64}a^{6}+a^{3}+a^{2}-16+\frac{1}{64}a^{6}-\frac{3}{16}a^{4}a+\frac{3}{4}a^{2}a^{2}-a^{3}-\left(3a-8\right)\left(3a+8\right)-3\left(\frac{1}{2}a^{2}-4\right)^{2}$$

To multiply powers of the same base, add their exponents. Add 4 and 1 to get 5.

$$-\frac{1}{64}a^{6}+a^{3}+a^{2}-16+\frac{1}{64}a^{6}-\frac{3}{16}a^{5}+\frac{3}{4}a^{2}a^{2}-a^{3}-\left(3a-8\right)\left(3a+8\right)-3\left(\frac{1}{2}a^{2}-4\right)^{2}$$

To multiply powers of the same base, add their exponents. Add 2 and 2 to get 4.

$$-\frac{1}{64}a^{6}+a^{3}+a^{2}-16+\frac{1}{64}a^{6}-\frac{3}{16}a^{5}+\frac{3}{4}a^{4}-a^{3}-\left(3a-8\right)\left(3a+8\right)-3\left(\frac{1}{2}a^{2}-4\right)^{2}$$

Combine -\frac{1}{64}a^{6} and \frac{1}{64}a^{6} to get 0.

$$a^{3}+a^{2}-16-\frac{3}{16}a^{5}+\frac{3}{4}a^{4}-a^{3}-\left(3a-8\right)\left(3a+8\right)-3\left(\frac{1}{2}a^{2}-4\right)^{2}$$

Combine a^{3} and -a^{3} to get 0.

$$a^{2}-16-\frac{3}{16}a^{5}+\frac{3}{4}a^{4}-\left(3a-8\right)\left(3a+8\right)-3\left(\frac{1}{2}a^{2}-4\right)^{2}$$

Consider \left(3a-8\right)\left(3a+8\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 8.

$$a^{2}-16-\frac{3}{16}a^{5}+\frac{3}{4}a^{4}-\left(\left(3a\right)^{2}-64\right)-3\left(\frac{1}{2}a^{2}-4\right)^{2}$$

Expand \left(3a\right)^{2}.

$$a^{2}-16-\frac{3}{16}a^{5}+\frac{3}{4}a^{4}-\left(3^{2}a^{2}-64\right)-3\left(\frac{1}{2}a^{2}-4\right)^{2}$$

Calculate 3 to the power of 2 and get 9.

$$a^{2}-16-\frac{3}{16}a^{5}+\frac{3}{4}a^{4}-\left(9a^{2}-64\right)-3\left(\frac{1}{2}a^{2}-4\right)^{2}$$

To find the opposite of 9a^{2}-64, find the opposite of each term.

$$a^{2}-16-\frac{3}{16}a^{5}+\frac{3}{4}a^{4}-9a^{2}+64-3\left(\frac{1}{2}a^{2}-4\right)^{2}$$

Combine a^{2} and -9a^{2} to get -8a^{2}.

$$-8a^{2}-16-\frac{3}{16}a^{5}+\frac{3}{4}a^{4}+64-3\left(\frac{1}{2}a^{2}-4\right)^{2}$$

Add -16 and 64 to get 48.

$$-8a^{2}+48-\frac{3}{16}a^{5}+\frac{3}{4}a^{4}-3\left(\frac{1}{2}a^{2}-4\right)^{2}$$

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

$$-8a^{2}+48-\frac{3}{16}a^{5}+\frac{3}{4}a^{4}-3\left(\frac{1}{4}\left(a^{2}\right)^{2}-4a^{2}+16\right)$$

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

$$-8a^{2}+48-\frac{3}{16}a^{5}+\frac{3}{4}a^{4}-3\left(\frac{1}{4}a^{4}-4a^{2}+16\right)$$

Use the distributive property to multiply -3 by \frac{1}{4}a^{4}-4a^{2}+16.

$$-8a^{2}+48-\frac{3}{16}a^{5}+\frac{3}{4}a^{4}-\frac{3}{4}a^{4}+12a^{2}-48$$

Combine \frac{3}{4}a^{4} and -\frac{3}{4}a^{4} to get 0.

$$-8a^{2}+48-\frac{3}{16}a^{5}+12a^{2}-48$$

Combine -8a^{2} and 12a^{2} to get 4a^{2}.

$$4a^{2}+48-\frac{3}{16}a^{5}-48$$

Subtract 48 from 48 to get 0.

$$4a^{2}-\frac{3}{16}a^{5}$$