Solve by Factoring a^2+2ab+b^2=(a+b)^2

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

Step 1

Subtract ${(a + b)}^{2}$ from both sides of the equation.

$a^{2} + 2 a b + b^{2} - {(a + b)}^{2} = 0$

Step 2

Simplify $a^{2} + 2 a b + b^{2} - {(a + b)}^{2}$ .

Simplify each term.

Rewrite ${(a + b)}^{2}$ as $(a + b) (a + b)$ .

$a^{2} + 2 a b + b^{2} - ((a + b) (a + b)) = 0$

Expand $(a + b) (a + b)$ using the FOIL Method.

Apply the distributive property.

$a^{2} + 2 a b + b^{2} - (a (a + b) + b (a + b)) = 0$

Apply the distributive property.

$a^{2} + 2 a b + b^{2} - (a \cdot a + a b + b (a + b)) = 0$

Apply the distributive property.

$a^{2} + 2 a b + b^{2} - (a \cdot a + a b + b a + b \cdot b) = 0$

Simplify and combine like terms.

Simplify each term.

Multiply $a$ by $a$ .

$a^{2} + 2 a b + b^{2} - (a^{2} + a b + b a + b \cdot b) = 0$

Multiply $b$ by $b$ .

$a^{2} + 2 a b + b^{2} - (a^{2} + a b + b a + b^{2}) = 0$

Add $a b$ and $b a$ .

Reorder $b$ and $a$ .

$a^{2} + 2 a b + b^{2} - (a^{2} + a b + a b + b^{2}) = 0$

Add $a b$ and $a b$ .

$a^{2} + 2 a b + b^{2} - (a^{2} + 2 a b + b^{2}) = 0$

Apply the distributive property.

$a^{2} + 2 a b + b^{2} - a^{2} - (2 a b) - b^{2} = 0$

Multiply $2$ by $- 1$ .

$a^{2} + 2 a b + b^{2} - a^{2} - 2 a b - b^{2} = 0$

Combine the opposite terms in $a^{2} + 2 a b + b^{2} - a^{2} - 2 a b - b^{2}$ .

Subtract $a^{2}$ from $a^{2}$ .

$2 a b + b^{2} + 0 - 2 a b - b^{2} = 0$

Add $2 a b + b^{2}$ and $0$ .

$2 a b + b^{2} - 2 a b - b^{2} = 0$

Subtract $2 a b$ from $2 a b$ .

$b^{2} + 0 - b^{2} = 0$

Add $b^{2}$ and $0$ .

$b^{2} - b^{2} = 0$

Subtract $b^{2}$ from $b^{2}$ .

$0 = 0$

Step 3

Since $0 = 0$ , the equation will always be true.

Always true