Solve (2a-3b)^2=(3b-2a)^2

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

Solve for a (complex solution)

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

Solution Steps

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

$$4a^{2}-12ab+9b^{2}=\left(3b-2a\right)^{2}$$

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

$$4a^{2}-12ab+9b^{2}=9b^{2}-12ba+4a^{2}$$

Add 12ba to both sides.

$$4a^{2}-12ab+9b^{2}+12ba=9b^{2}+4a^{2}$$

Combine -12ab and 12ba to get 0.

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

Subtract 4a^{2} from both sides.

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

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

$$9b^{2}=9b^{2}$$

Cancel out 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

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

$$4a^{2}-12ab+9b^{2}=\left(3b-2a\right)^{2}$$

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

$$4a^{2}-12ab+9b^{2}=9b^{2}-12ba+4a^{2}$$

Subtract 9b^{2} from both sides.

$$4a^{2}-12ab+9b^{2}-9b^{2}=-12ba+4a^{2}$$

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

$$4a^{2}-12ab=-12ba+4a^{2}$$

Add 12ba to both sides.

$$4a^{2}-12ab+12ba=4a^{2}$$

Combine -12ab and 12ba to get 0.

$$4a^{2}=4a^{2}$$

Cancel out 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

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

$$4a^{2}-12ab+9b^{2}=\left(3b-2a\right)^{2}$$

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

$$4a^{2}-12ab+9b^{2}=9b^{2}-12ba+4a^{2}$$

Add 12ba to both sides.

$$4a^{2}-12ab+9b^{2}+12ba=9b^{2}+4a^{2}$$

Combine -12ab and 12ba to get 0.

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

Subtract 4a^{2} from both sides.

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

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

$$9b^{2}=9b^{2}$$

Cancel out 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

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

$$4a^{2}-12ab+9b^{2}=\left(3b-2a\right)^{2}$$

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

$$4a^{2}-12ab+9b^{2}=9b^{2}-12ba+4a^{2}$$

Subtract 9b^{2} from both sides.

$$4a^{2}-12ab+9b^{2}-9b^{2}=-12ba+4a^{2}$$

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

$$4a^{2}-12ab=-12ba+4a^{2}$$

Add 12ba to both sides.

$$4a^{2}-12ab+12ba=4a^{2}$$

Combine -12ab and 12ba to get 0.

$$4a^{2}=4a^{2}$$

Cancel out 4 on both sides.

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

Reorder the terms.

$$\text{true}$$

This is true for any b.

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