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

$$( a + b ^ { 2 } ) ^ { 3 } - b ^ { 4 } [ 3 ( a - \frac { 1 } { 2 } ) + ( b + 1 ) ( b - 1 ) ] + ( - \frac { 3 } { 2 } ) ( a ^ { 2 } + b ^ { 2 } ) ^ { 2 }$$

Evaluate

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

Solution Steps

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

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

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

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

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

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

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

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

Consider \left(b+1\right)\left(b-1\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.

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

Subtract 1 from -\frac{3}{2} to get -\frac{5}{2}.

$$a^{3}+3a^{2}b^{2}+3ab^{4}+b^{6}-b^{4}\left(3a-\frac{5}{2}+b^{2}\right)-\frac{3}{2}\left(a^{2}+b^{2}\right)^{2}$$

Use the distributive property to multiply b^{4} by 3a-\frac{5}{2}+b^{2}.

$$a^{3}+3a^{2}b^{2}+3ab^{4}+b^{6}-\left(3b^{4}a-\frac{5}{2}b^{4}+b^{6}\right)-\frac{3}{2}\left(a^{2}+b^{2}\right)^{2}$$

To find the opposite of 3b^{4}a-\frac{5}{2}b^{4}+b^{6}, find the opposite of each term.

$$a^{3}+3a^{2}b^{2}+3ab^{4}+b^{6}-3b^{4}a+\frac{5}{2}b^{4}-b^{6}-\frac{3}{2}\left(a^{2}+b^{2}\right)^{2}$$

Combine 3ab^{4} and -3b^{4}a to get 0.

$$a^{3}+3a^{2}b^{2}+b^{6}+\frac{5}{2}b^{4}-b^{6}-\frac{3}{2}\left(a^{2}+b^{2}\right)^{2}$$

Combine b^{6} and -b^{6} to get 0.

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

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

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

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

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

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

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

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

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

Combine 3a^{2}b^{2} and -3a^{2}b^{2} to get 0.

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

Combine \frac{5}{2}b^{4} and -\frac{3}{2}b^{4} to get b^{4}.

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

Expand

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

Solution Steps

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

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

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

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

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

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

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

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

Consider \left(b+1\right)\left(b-1\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.

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

Subtract 1 from -\frac{3}{2} to get -\frac{5}{2}.

$$a^{3}+3a^{2}b^{2}+3ab^{4}+b^{6}-b^{4}\left(3a-\frac{5}{2}+b^{2}\right)-\frac{3}{2}\left(a^{2}+b^{2}\right)^{2}$$

Use the distributive property to multiply b^{4} by 3a-\frac{5}{2}+b^{2}.

$$a^{3}+3a^{2}b^{2}+3ab^{4}+b^{6}-\left(3b^{4}a-\frac{5}{2}b^{4}+b^{6}\right)-\frac{3}{2}\left(a^{2}+b^{2}\right)^{2}$$

To find the opposite of 3b^{4}a-\frac{5}{2}b^{4}+b^{6}, find the opposite of each term.

$$a^{3}+3a^{2}b^{2}+3ab^{4}+b^{6}-3b^{4}a+\frac{5}{2}b^{4}-b^{6}-\frac{3}{2}\left(a^{2}+b^{2}\right)^{2}$$

Combine 3ab^{4} and -3b^{4}a to get 0.

$$a^{3}+3a^{2}b^{2}+b^{6}+\frac{5}{2}b^{4}-b^{6}-\frac{3}{2}\left(a^{2}+b^{2}\right)^{2}$$

Combine b^{6} and -b^{6} to get 0.

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

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

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

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

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

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

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

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

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

Combine 3a^{2}b^{2} and -3a^{2}b^{2} to get 0.

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

Combine \frac{5}{2}b^{4} and -\frac{3}{2}b^{4} to get b^{4}.

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