Solve (3/2a-2/3b)(3/2a-2/3b)=(frac{3}{2}a-frac{2}{3}b)^2

$$( \frac { 3 } { 2 } a - \frac { 2 } { 3 } b ) ( \frac { 3 } { 2 } a - \frac { 2 } { 3 } b ) = ( \frac { 3 } { 2 } a - \frac { 2 } { 3 } b ) ^ { 2 }$$

Solve for a (complex solution)

$$a\in \mathrm{C}$$

Solution Steps

Multiply \frac{3}{2}a-\frac{2}{3}b and \frac{3}{2}a-\frac{2}{3}b to get \left(\frac{3}{2}a-\frac{2}{3}b\right)^{2}.

$$\left(\frac{3}{2}a-\frac{2}{3}b\right)^{2}=\left(\frac{3}{2}a-\frac{2}{3}b\right)^{2}$$

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

$$\frac{9}{4}a^{2}-2ab+\frac{4}{9}b^{2}=\left(\frac{3}{2}a-\frac{2}{3}b\right)^{2}$$

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

$$\frac{9}{4}a^{2}-2ab+\frac{4}{9}b^{2}=\frac{9}{4}a^{2}-2ab+\frac{4}{9}b^{2}$$

Subtract \frac{9}{4}a^{2} from both sides.

$$\frac{9}{4}a^{2}-2ab+\frac{4}{9}b^{2}-\frac{9}{4}a^{2}=-2ab+\frac{4}{9}b^{2}$$

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

$$-2ab+\frac{4}{9}b^{2}=-2ab+\frac{4}{9}b^{2}$$

Add 2ab to both sides.

$$-2ab+\frac{4}{9}b^{2}+2ab=\frac{4}{9}b^{2}$$

Combine -2ab and 2ab to get 0.

$$\frac{4}{9}b^{2}=\frac{4}{9}b^{2}$$

Cancel out \frac{4}{9} on both sides.

$$b^{2}=b^{2}$$

Reorder the terms.

$$\text{true}$$

This is true for any a.

$$a\in \mathrm{C}$$

Solve for b (complex solution)

$$b\in \mathrm{C}$$

Solution Steps

Multiply \frac{3}{2}a-\frac{2}{3}b and \frac{3}{2}a-\frac{2}{3}b to get \left(\frac{3}{2}a-\frac{2}{3}b\right)^{2}.

$$\left(\frac{3}{2}a-\frac{2}{3}b\right)^{2}=\left(\frac{3}{2}a-\frac{2}{3}b\right)^{2}$$

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

$$\frac{9}{4}a^{2}-2ab+\frac{4}{9}b^{2}=\left(\frac{3}{2}a-\frac{2}{3}b\right)^{2}$$

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

$$\frac{9}{4}a^{2}-2ab+\frac{4}{9}b^{2}=\frac{9}{4}a^{2}-2ab+\frac{4}{9}b^{2}$$

Add 2ab to both sides.

$$\frac{9}{4}a^{2}-2ab+\frac{4}{9}b^{2}+2ab=\frac{9}{4}a^{2}+\frac{4}{9}b^{2}$$

Combine -2ab and 2ab to get 0.

$$\frac{9}{4}a^{2}+\frac{4}{9}b^{2}=\frac{9}{4}a^{2}+\frac{4}{9}b^{2}$$

Subtract \frac{4}{9}b^{2} from both sides.

$$\frac{9}{4}a^{2}+\frac{4}{9}b^{2}-\frac{4}{9}b^{2}=\frac{9}{4}a^{2}$$

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

$$\frac{9}{4}a^{2}=\frac{9}{4}a^{2}$$

Cancel out \frac{9}{4} on both sides.

$$a^{2}=a^{2}$$

Reorder the terms.

$$\text{true}$$

This is true for any b.

$$b\in \mathrm{C}$$

Solve for a

$$a\in \mathrm{R}$$

Solution Steps

Multiply \frac{3}{2}a-\frac{2}{3}b and \frac{3}{2}a-\frac{2}{3}b to get \left(\frac{3}{2}a-\frac{2}{3}b\right)^{2}.

$$\left(\frac{3}{2}a-\frac{2}{3}b\right)^{2}=\left(\frac{3}{2}a-\frac{2}{3}b\right)^{2}$$

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

$$\frac{9}{4}a^{2}-2ab+\frac{4}{9}b^{2}=\left(\frac{3}{2}a-\frac{2}{3}b\right)^{2}$$

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

$$\frac{9}{4}a^{2}-2ab+\frac{4}{9}b^{2}=\frac{9}{4}a^{2}-2ab+\frac{4}{9}b^{2}$$

Subtract \frac{9}{4}a^{2} from both sides.

$$\frac{9}{4}a^{2}-2ab+\frac{4}{9}b^{2}-\frac{9}{4}a^{2}=-2ab+\frac{4}{9}b^{2}$$

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

$$-2ab+\frac{4}{9}b^{2}=-2ab+\frac{4}{9}b^{2}$$

Add 2ab to both sides.

$$-2ab+\frac{4}{9}b^{2}+2ab=\frac{4}{9}b^{2}$$

Combine -2ab and 2ab to get 0.

$$\frac{4}{9}b^{2}=\frac{4}{9}b^{2}$$

Cancel out \frac{4}{9} on both sides.

$$b^{2}=b^{2}$$

Reorder the terms.

$$\text{true}$$

This is true for any a.

$$a\in \mathrm{R}$$

Solve for b

$$b\in \mathrm{R}$$

Solution Steps

Multiply \frac{3}{2}a-\frac{2}{3}b and \frac{3}{2}a-\frac{2}{3}b to get \left(\frac{3}{2}a-\frac{2}{3}b\right)^{2}.

$$\left(\frac{3}{2}a-\frac{2}{3}b\right)^{2}=\left(\frac{3}{2}a-\frac{2}{3}b\right)^{2}$$

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

$$\frac{9}{4}a^{2}-2ab+\frac{4}{9}b^{2}=\left(\frac{3}{2}a-\frac{2}{3}b\right)^{2}$$

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

$$\frac{9}{4}a^{2}-2ab+\frac{4}{9}b^{2}=\frac{9}{4}a^{2}-2ab+\frac{4}{9}b^{2}$$

Add 2ab to both sides.

$$\frac{9}{4}a^{2}-2ab+\frac{4}{9}b^{2}+2ab=\frac{9}{4}a^{2}+\frac{4}{9}b^{2}$$

Combine -2ab and 2ab to get 0.

$$\frac{9}{4}a^{2}+\frac{4}{9}b^{2}=\frac{9}{4}a^{2}+\frac{4}{9}b^{2}$$

Subtract \frac{4}{9}b^{2} from both sides.

$$\frac{9}{4}a^{2}+\frac{4}{9}b^{2}-\frac{4}{9}b^{2}=\frac{9}{4}a^{2}$$

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

$$\frac{9}{4}a^{2}=\frac{9}{4}a^{2}$$

Cancel out \frac{9}{4} on both sides.

$$a^{2}=a^{2}$$

Reorder the terms.

$$\text{true}$$

This is true for any b.

$$b\in \mathrm{R}$$