Solve left(a-bright)*left(2a^2-1/5ab+b^2right)=2a^3-11/5a^2b+6/5a{b}^2-{b}^3

$$\left( a-b \right) \times \left( 2 { a }^{ 2 } - \frac{ 1 }{ 5 } ab+ { b }^{ 2 } \right) =2 { a }^{ 3 } - \frac{ 11 }{ 5 } { a }^{ 2 } b+ \frac{ 6 }{ 5 } a { b }^{ 2 } - { b }^{ 3 }$$

Solve for a (complex solution)

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

Solution Steps

Use the distributive property to multiply a-b by 2a^{2}-\frac{1}{5}ab+b^{2} and combine like terms.

$$2a^{3}-\frac{11}{5}ba^{2}+\frac{6}{5}ab^{2}-b^{3}=2a^{3}-\frac{11}{5}a^{2}b+\frac{6}{5}ab^{2}-b^{3}$$

Subtract 2a^{3}-\frac{11}{5}a^{2}b from both sides.

$$2a^{3}-\frac{11}{5}ba^{2}+\frac{6}{5}ab^{2}-b^{3}-\left(2a^{3}-\frac{11}{5}a^{2}b\right)=\frac{6}{5}ab^{2}-b^{3}$$

Subtract \frac{6}{5}ab^{2} from both sides.

$$2a^{3}-\frac{11}{5}ba^{2}+\frac{6}{5}ab^{2}-b^{3}-\left(2a^{3}-\frac{11}{5}a^{2}b\right)-\frac{6}{5}ab^{2}=-b^{3}$$

To find the opposite of 2a^{3}-\frac{11}{5}a^{2}b, find the opposite of each term.

$$2a^{3}-\frac{11}{5}ba^{2}+\frac{6}{5}ab^{2}-b^{3}-2a^{3}+\frac{11}{5}a^{2}b-\frac{6}{5}ab^{2}=-b^{3}$$

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

$$-\frac{11}{5}ba^{2}+\frac{6}{5}ab^{2}-b^{3}+\frac{11}{5}a^{2}b-\frac{6}{5}ab^{2}=-b^{3}$$

Combine -\frac{11}{5}ba^{2} and \frac{11}{5}a^{2}b to get 0.

$$\frac{6}{5}ab^{2}-b^{3}-\frac{6}{5}ab^{2}=-b^{3}$$

Combine \frac{6}{5}ab^{2} and -\frac{6}{5}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 2a^{2}-\frac{1}{5}ab+b^{2} and combine like terms.

$$2a^{3}-\frac{11}{5}ba^{2}+\frac{6}{5}ab^{2}-b^{3}=2a^{3}-\frac{11}{5}a^{2}b+\frac{6}{5}ab^{2}-b^{3}$$

Subtract 2a^{3}-\frac{11}{5}a^{2}b from both sides.

$$2a^{3}-\frac{11}{5}ba^{2}+\frac{6}{5}ab^{2}-b^{3}-\left(2a^{3}-\frac{11}{5}a^{2}b\right)=\frac{6}{5}ab^{2}-b^{3}$$

Subtract \frac{6}{5}ab^{2} from both sides.

$$2a^{3}-\frac{11}{5}ba^{2}+\frac{6}{5}ab^{2}-b^{3}-\left(2a^{3}-\frac{11}{5}a^{2}b\right)-\frac{6}{5}ab^{2}=-b^{3}$$

Add b^{3} to both sides.

$$2a^{3}-\frac{11}{5}ba^{2}+\frac{6}{5}ab^{2}-b^{3}-\left(2a^{3}-\frac{11}{5}a^{2}b\right)-\frac{6}{5}ab^{2}+b^{3}=0$$

To find the opposite of 2a^{3}-\frac{11}{5}a^{2}b, find the opposite of each term.

$$2a^{3}-\frac{11}{5}ba^{2}+\frac{6}{5}ab^{2}-b^{3}-2a^{3}+\frac{11}{5}a^{2}b-\frac{6}{5}ab^{2}+b^{3}=0$$

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

$$-\frac{11}{5}ba^{2}+\frac{6}{5}ab^{2}-b^{3}+\frac{11}{5}a^{2}b-\frac{6}{5}ab^{2}+b^{3}=0$$

Combine -\frac{11}{5}ba^{2} and \frac{11}{5}a^{2}b to get 0.

$$\frac{6}{5}ab^{2}-b^{3}-\frac{6}{5}ab^{2}+b^{3}=0$$

Combine \frac{6}{5}ab^{2} and -\frac{6}{5}ab^{2} to get 0.

$$-b^{3}+b^{3}=0$$

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

$$0=0$$

Compare 0 and 0.

$$\text{true}$$

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 2a^{2}-\frac{1}{5}ab+b^{2} and combine like terms.

$$2a^{3}-\frac{11}{5}ba^{2}+\frac{6}{5}ab^{2}-b^{3}=2a^{3}-\frac{11}{5}a^{2}b+\frac{6}{5}ab^{2}-b^{3}$$

Subtract 2a^{3}-\frac{11}{5}a^{2}b from both sides.

$$2a^{3}-\frac{11}{5}ba^{2}+\frac{6}{5}ab^{2}-b^{3}-\left(2a^{3}-\frac{11}{5}a^{2}b\right)=\frac{6}{5}ab^{2}-b^{3}$$

Subtract \frac{6}{5}ab^{2} from both sides.

$$2a^{3}-\frac{11}{5}ba^{2}+\frac{6}{5}ab^{2}-b^{3}-\left(2a^{3}-\frac{11}{5}a^{2}b\right)-\frac{6}{5}ab^{2}=-b^{3}$$

To find the opposite of 2a^{3}-\frac{11}{5}a^{2}b, find the opposite of each term.

$$2a^{3}-\frac{11}{5}ba^{2}+\frac{6}{5}ab^{2}-b^{3}-2a^{3}+\frac{11}{5}a^{2}b-\frac{6}{5}ab^{2}=-b^{3}$$

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

$$-\frac{11}{5}ba^{2}+\frac{6}{5}ab^{2}-b^{3}+\frac{11}{5}a^{2}b-\frac{6}{5}ab^{2}=-b^{3}$$

Combine -\frac{11}{5}ba^{2} and \frac{11}{5}a^{2}b to get 0.

$$\frac{6}{5}ab^{2}-b^{3}-\frac{6}{5}ab^{2}=-b^{3}$$

Combine \frac{6}{5}ab^{2} and -\frac{6}{5}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 2a^{2}-\frac{1}{5}ab+b^{2} and combine like terms.

$$2a^{3}-\frac{11}{5}ba^{2}+\frac{6}{5}ab^{2}-b^{3}=2a^{3}-\frac{11}{5}a^{2}b+\frac{6}{5}ab^{2}-b^{3}$$

Subtract 2a^{3}-\frac{11}{5}a^{2}b from both sides.

$$2a^{3}-\frac{11}{5}ba^{2}+\frac{6}{5}ab^{2}-b^{3}-\left(2a^{3}-\frac{11}{5}a^{2}b\right)=\frac{6}{5}ab^{2}-b^{3}$$

Subtract \frac{6}{5}ab^{2} from both sides.

$$2a^{3}-\frac{11}{5}ba^{2}+\frac{6}{5}ab^{2}-b^{3}-\left(2a^{3}-\frac{11}{5}a^{2}b\right)-\frac{6}{5}ab^{2}=-b^{3}$$

Add b^{3} to both sides.

$$2a^{3}-\frac{11}{5}ba^{2}+\frac{6}{5}ab^{2}-b^{3}-\left(2a^{3}-\frac{11}{5}a^{2}b\right)-\frac{6}{5}ab^{2}+b^{3}=0$$

To find the opposite of 2a^{3}-\frac{11}{5}a^{2}b, find the opposite of each term.

$$2a^{3}-\frac{11}{5}ba^{2}+\frac{6}{5}ab^{2}-b^{3}-2a^{3}+\frac{11}{5}a^{2}b-\frac{6}{5}ab^{2}+b^{3}=0$$

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

$$-\frac{11}{5}ba^{2}+\frac{6}{5}ab^{2}-b^{3}+\frac{11}{5}a^{2}b-\frac{6}{5}ab^{2}+b^{3}=0$$

Combine -\frac{11}{5}ba^{2} and \frac{11}{5}a^{2}b to get 0.

$$\frac{6}{5}ab^{2}-b^{3}-\frac{6}{5}ab^{2}+b^{3}=0$$

Combine \frac{6}{5}ab^{2} and -\frac{6}{5}ab^{2} to get 0.

$$-b^{3}+b^{3}=0$$

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

$$0=0$$

Compare 0 and 0.

$$\text{true}$$

This is true for any b.

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