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

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

Solve for a (complex solution)

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

Solution Steps

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

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

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

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

Use the distributive property to multiply a-b by a^{2}+2ab+b^{2} and combine like terms.

$$a^{3}-ab^{2}+ba^{2}-b^{3}=a^{3}+ba^{2}-ab^{2}-b^{3}$$

Subtract a^{3} from both sides.

$$a^{3}-ab^{2}+ba^{2}-b^{3}-a^{3}=ba^{2}-ab^{2}-b^{3}$$

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

$$-ab^{2}+ba^{2}-b^{3}=ba^{2}-ab^{2}-b^{3}$$

Subtract ba^{2} from both sides.

$$-ab^{2}+ba^{2}-b^{3}-ba^{2}=-ab^{2}-b^{3}$$

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

$$-ab^{2}-b^{3}=-ab^{2}-b^{3}$$

Add ab^{2} to both sides.

$$-ab^{2}-b^{3}+ab^{2}=-b^{3}$$

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

$$-b^{3}=-b^{3}$$

Cancel out -1 on both sides.

$$b^{3}=b^{3}$$

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 the distributive property to multiply a+b by a^{2}-b^{2}.

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

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

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

Use the distributive property to multiply a-b by a^{2}+2ab+b^{2} and combine like terms.

$$a^{3}-ab^{2}+ba^{2}-b^{3}=a^{3}+ba^{2}-ab^{2}-b^{3}$$

Subtract ba^{2} from both sides.

$$a^{3}-ab^{2}+ba^{2}-b^{3}-ba^{2}=a^{3}-ab^{2}-b^{3}$$

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

$$a^{3}-ab^{2}-b^{3}=a^{3}-ab^{2}-b^{3}$$

Add ab^{2} to both sides.

$$a^{3}-ab^{2}-b^{3}+ab^{2}=a^{3}-b^{3}$$

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

$$a^{3}-b^{3}=a^{3}-b^{3}$$

Add b^{3} to both sides.

$$a^{3}-b^{3}+b^{3}=a^{3}$$

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

$$a^{3}=a^{3}$$

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 the distributive property to multiply a+b by a^{2}-b^{2}.

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

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

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

Use the distributive property to multiply a-b by a^{2}+2ab+b^{2} and combine like terms.

$$a^{3}-ab^{2}+ba^{2}-b^{3}=a^{3}+ba^{2}-ab^{2}-b^{3}$$

Subtract a^{3} from both sides.

$$a^{3}-ab^{2}+ba^{2}-b^{3}-a^{3}=ba^{2}-ab^{2}-b^{3}$$

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

$$-ab^{2}+ba^{2}-b^{3}=ba^{2}-ab^{2}-b^{3}$$

Subtract ba^{2} from both sides.

$$-ab^{2}+ba^{2}-b^{3}-ba^{2}=-ab^{2}-b^{3}$$

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

$$-ab^{2}-b^{3}=-ab^{2}-b^{3}$$

Add ab^{2} to both sides.

$$-ab^{2}-b^{3}+ab^{2}=-b^{3}$$

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

$$-b^{3}=-b^{3}$$

Cancel out -1 on both sides.

$$b^{3}=b^{3}$$

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 the distributive property to multiply a+b by a^{2}-b^{2}.

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

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

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

Use the distributive property to multiply a-b by a^{2}+2ab+b^{2} and combine like terms.

$$a^{3}-ab^{2}+ba^{2}-b^{3}=a^{3}+ba^{2}-ab^{2}-b^{3}$$

Subtract ba^{2} from both sides.

$$a^{3}-ab^{2}+ba^{2}-b^{3}-ba^{2}=a^{3}-ab^{2}-b^{3}$$

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

$$a^{3}-ab^{2}-b^{3}=a^{3}-ab^{2}-b^{3}$$

Add ab^{2} to both sides.

$$a^{3}-ab^{2}-b^{3}+ab^{2}=a^{3}-b^{3}$$

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

$$a^{3}-b^{3}=a^{3}-b^{3}$$

Add b^{3} to both sides.

$$a^{3}-b^{3}+b^{3}=a^{3}$$

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

$$a^{3}=a^{3}$$

Reorder the terms.

$$\text{true}$$

This is true for any b.

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