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

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

Evaluate

$$4\left(ac\right)^{2}$$

Solution Steps

Square \frac{1}{2}a+\frac{2}{3}b+c.

$$\left(\frac{1}{4}a^{2}+\frac{2}{3}ab+ac+\frac{4}{9}b^{2}+\frac{4}{3}bc+c^{2}-\left(\frac{1}{2}a+\frac{2}{3}b-c\right)^{2}+\frac{1}{3}bc\right)^{2}-3bc^{2}\left(4a+3b\right)$$

Square \frac{1}{2}a+\frac{2}{3}b-c.

$$\left(\frac{1}{4}a^{2}+\frac{2}{3}ab+ac+\frac{4}{9}b^{2}+\frac{4}{3}bc+c^{2}-\left(\frac{1}{4}a^{2}+\frac{2}{3}ab-ac+\frac{4}{9}b^{2}-\frac{4}{3}bc+c^{2}\right)+\frac{1}{3}bc\right)^{2}-3bc^{2}\left(4a+3b\right)$$

To find the opposite of \frac{1}{4}a^{2}+\frac{2}{3}ab-ac+\frac{4}{9}b^{2}-\frac{4}{3}bc+c^{2}, find the opposite of each term.

$$\left(\frac{1}{4}a^{2}+\frac{2}{3}ab+ac+\frac{4}{9}b^{2}+\frac{4}{3}bc+c^{2}-\frac{1}{4}a^{2}-\frac{2}{3}ab+ac-\frac{4}{9}b^{2}+\frac{4}{3}bc-c^{2}+\frac{1}{3}bc\right)^{2}-3bc^{2}\left(4a+3b\right)$$

Combine \frac{1}{4}a^{2} and -\frac{1}{4}a^{2} to get 0.

$$\left(\frac{2}{3}ab+ac+\frac{4}{9}b^{2}+\frac{4}{3}bc+c^{2}-\frac{2}{3}ab+ac-\frac{4}{9}b^{2}+\frac{4}{3}bc-c^{2}+\frac{1}{3}bc\right)^{2}-3bc^{2}\left(4a+3b\right)$$

Combine \frac{2}{3}ab and -\frac{2}{3}ab to get 0.

$$\left(ac+\frac{4}{9}b^{2}+\frac{4}{3}bc+c^{2}+ac-\frac{4}{9}b^{2}+\frac{4}{3}bc-c^{2}+\frac{1}{3}bc\right)^{2}-3bc^{2}\left(4a+3b\right)$$

Combine ac and ac to get 2ac.

$$\left(2ac+\frac{4}{9}b^{2}+\frac{4}{3}bc+c^{2}-\frac{4}{9}b^{2}+\frac{4}{3}bc-c^{2}+\frac{1}{3}bc\right)^{2}-3bc^{2}\left(4a+3b\right)$$

Combine \frac{4}{9}b^{2} and -\frac{4}{9}b^{2} to get 0.

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

Combine \frac{4}{3}bc and \frac{4}{3}bc to get \frac{8}{3}bc.

$$\left(2ac+\frac{8}{3}bc+c^{2}-c^{2}+\frac{1}{3}bc\right)^{2}-3bc^{2}\left(4a+3b\right)$$

Combine c^{2} and -c^{2} to get 0.

$$\left(2ac+\frac{8}{3}bc+\frac{1}{3}bc\right)^{2}-3bc^{2}\left(4a+3b\right)$$

Combine \frac{8}{3}bc and \frac{1}{3}bc to get 3bc.

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

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

$$4a^{2}c^{2}+12acbc+9b^{2}c^{2}-3bc^{2}\left(4a+3b\right)$$

Multiply c and c to get c^{2}.

$$4a^{2}c^{2}+12ac^{2}b+9b^{2}c^{2}-3bc^{2}\left(4a+3b\right)$$

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

$$4a^{2}c^{2}+12ac^{2}b+9b^{2}c^{2}-12bc^{2}a-9b^{2}c^{2}$$

Combine 12ac^{2}b and -12bc^{2}a to get 0.

$$4a^{2}c^{2}+9b^{2}c^{2}-9b^{2}c^{2}$$

Combine 9b^{2}c^{2} and -9b^{2}c^{2} to get 0.

$$4a^{2}c^{2}$$

Expand

$$4\left(ac\right)^{2}$$

Solution Steps

Square \frac{1}{2}a+\frac{2}{3}b+c.

$$\left(\frac{1}{4}a^{2}+\frac{2}{3}ab+ac+\frac{4}{9}b^{2}+\frac{4}{3}bc+c^{2}-\left(\frac{1}{2}a+\frac{2}{3}b-c\right)^{2}+\frac{1}{3}bc\right)^{2}-3bc^{2}\left(4a+3b\right)$$

Square \frac{1}{2}a+\frac{2}{3}b-c.

$$\left(\frac{1}{4}a^{2}+\frac{2}{3}ab+ac+\frac{4}{9}b^{2}+\frac{4}{3}bc+c^{2}-\left(\frac{1}{4}a^{2}+\frac{2}{3}ab-ac+\frac{4}{9}b^{2}-\frac{4}{3}bc+c^{2}\right)+\frac{1}{3}bc\right)^{2}-3bc^{2}\left(4a+3b\right)$$

To find the opposite of \frac{1}{4}a^{2}+\frac{2}{3}ab-ac+\frac{4}{9}b^{2}-\frac{4}{3}bc+c^{2}, find the opposite of each term.

$$\left(\frac{1}{4}a^{2}+\frac{2}{3}ab+ac+\frac{4}{9}b^{2}+\frac{4}{3}bc+c^{2}-\frac{1}{4}a^{2}-\frac{2}{3}ab+ac-\frac{4}{9}b^{2}+\frac{4}{3}bc-c^{2}+\frac{1}{3}bc\right)^{2}-3bc^{2}\left(4a+3b\right)$$

Combine \frac{1}{4}a^{2} and -\frac{1}{4}a^{2} to get 0.

$$\left(\frac{2}{3}ab+ac+\frac{4}{9}b^{2}+\frac{4}{3}bc+c^{2}-\frac{2}{3}ab+ac-\frac{4}{9}b^{2}+\frac{4}{3}bc-c^{2}+\frac{1}{3}bc\right)^{2}-3bc^{2}\left(4a+3b\right)$$

Combine \frac{2}{3}ab and -\frac{2}{3}ab to get 0.

$$\left(ac+\frac{4}{9}b^{2}+\frac{4}{3}bc+c^{2}+ac-\frac{4}{9}b^{2}+\frac{4}{3}bc-c^{2}+\frac{1}{3}bc\right)^{2}-3bc^{2}\left(4a+3b\right)$$

Combine ac and ac to get 2ac.

$$\left(2ac+\frac{4}{9}b^{2}+\frac{4}{3}bc+c^{2}-\frac{4}{9}b^{2}+\frac{4}{3}bc-c^{2}+\frac{1}{3}bc\right)^{2}-3bc^{2}\left(4a+3b\right)$$

Combine \frac{4}{9}b^{2} and -\frac{4}{9}b^{2} to get 0.

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

Combine \frac{4}{3}bc and \frac{4}{3}bc to get \frac{8}{3}bc.

$$\left(2ac+\frac{8}{3}bc+c^{2}-c^{2}+\frac{1}{3}bc\right)^{2}-3bc^{2}\left(4a+3b\right)$$

Combine c^{2} and -c^{2} to get 0.

$$\left(2ac+\frac{8}{3}bc+\frac{1}{3}bc\right)^{2}-3bc^{2}\left(4a+3b\right)$$

Combine \frac{8}{3}bc and \frac{1}{3}bc to get 3bc.

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

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

$$4a^{2}c^{2}+12acbc+9b^{2}c^{2}-3bc^{2}\left(4a+3b\right)$$

Multiply c and c to get c^{2}.

$$4a^{2}c^{2}+12ac^{2}b+9b^{2}c^{2}-3bc^{2}\left(4a+3b\right)$$

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

$$4a^{2}c^{2}+12ac^{2}b+9b^{2}c^{2}-12bc^{2}a-9b^{2}c^{2}$$

Combine 12ac^{2}b and -12bc^{2}a to get 0.

$$4a^{2}c^{2}+9b^{2}c^{2}-9b^{2}c^{2}$$

Combine 9b^{2}c^{2} and -9b^{2}c^{2} to get 0.

$$4a^{2}c^{2}$$