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

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

Solve for a (complex solution)

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

Solution Steps

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

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

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

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

To find the opposite of b^{3}-3b^{2}a+3ba^{2}-a^{3}, find the opposite of each term.

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

Subtract 3b^{2}a from both sides.

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

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

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

Add 3ba^{2} to both sides.

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

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

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

Subtract a^{3} from both sides.

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

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

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

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

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

To find the opposite of b^{3}-3b^{2}a+3ba^{2}-a^{3}, find the opposite of each term.

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

Add b^{3} to both sides.

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

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

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

Subtract 3b^{2}a from both sides.

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

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

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

Add 3ba^{2} to both sides.

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

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

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

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

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

To find the opposite of b^{3}-3b^{2}a+3ba^{2}-a^{3}, find the opposite of each term.

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

Subtract 3b^{2}a from both sides.

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

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

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

Add 3ba^{2} to both sides.

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

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

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

Subtract a^{3} from both sides.

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

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

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

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

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

To find the opposite of b^{3}-3b^{2}a+3ba^{2}-a^{3}, find the opposite of each term.

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

Add b^{3} to both sides.

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

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

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

Subtract 3b^{2}a from both sides.

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

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

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

Add 3ba^{2} to both sides.

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

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

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

Reorder the terms.

$$\text{true}$$

This is true for any b.

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