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

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

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}.

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

Add 3ab\left(a+b\right) to both sides.

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

Use the distributive property to multiply 3ab by a+b.

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

Subtract a^{3} from both sides.

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

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

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

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

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

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

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

Subtract 3ab^{2} from both sides.

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

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

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

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}.

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

Add 3ab\left(a+b\right) to both sides.

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

Use the distributive property to multiply 3ab by a+b.

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

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

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

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

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

Subtract 3ab^{2} from both sides.

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

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

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

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

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

Add 3ab\left(a+b\right) to both sides.

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

Use the distributive property to multiply 3ab by a+b.

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

Subtract a^{3} from both sides.

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

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

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

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

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

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

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

Subtract 3ab^{2} from both sides.

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

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

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

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}.

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

Add 3ab\left(a+b\right) to both sides.

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

Use the distributive property to multiply 3ab by a+b.

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

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

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

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

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

Subtract 3ab^{2} from both sides.

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

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

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

Subtract b^{3} from 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.

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