
An equation that is true for every value of the variable is called an identity. 
To show that an equation is an identity:
Start with either side of the equation and show that it can algebraically be changed into the other side. Or start with both sides of the equation and show that they both can be changed into the same algebraic expression.
You may be asked to verify (show, prove) that a known "rule" is an identity (always true). You may, or may not, be asked to supply "justification" for your work.
1. Show, with justification, that (x^{2} + y^{2})^{2} = (x^{2}  y^{2})^{2} + (2xy)^{2} is an identity.
x^{4} + 2x^{2}y^{2} +y^{4} = x^{4}  2x^{2}y^{2} +y^{4} + (4x^{2}y^{2}) 
expand by multiplication and distributive property. 
x^{4} + 2x^{2}y^{2} +y^{4} = x^{4} + 2x^{2}y^{2} +y^{4} 
combine like terms. 
(x^{2} + y^{2})^{2} = (x^{2}  y^{2})^{2} + (2xy)^{2} 
Identity! (always true) 
This identity is used to create Pythagorean Triples. For example, substituting x = 3 and y = 2 will give
13^{2} = 5^{2} + 12^{2} which is the Pythagorean Triple 5, 12, 13. 

You may be asked to verify (show, prove) that any algebraic equation is an identity.
NOTE: If you are asked to show that an equation is an identity, you may assume that the equation IS an identity. If you are asked to determine whether an equation is an identity, be careful as the answer may be that it is not an identity.
2. Show that
x^{3 }+ y^{3} = (x + y)(x^{2}  xy + y^{2})
is an identity.
Starting with the right hand side:
(distribute) (x + y)(x^{2}  xy + y^{2})
= x^{3}  x^{2}y + xy^{2} + x^{2}y  xy^{2} + y^{3}
(simplify) = x^{3 }+ y^{3}
This is an identity.
We have just shown that the formula for factoring the sum of two cubes is true. 
3. Determine whether
(2a^{2} + b^{3})(2a^{2}  b^{3}) = 4a^{2}  b^{3}
is an identity.
Starting with the left hand side:
(2a^{2} + b^{3})(2a^{2}  b^{3})
= 4a^{4} + 2a^{2}b^{3}  2a^{2}b^{3}  b^{6}
= 4a^{4}  b^{6}
4a^{4}  b^{6} ≠ 4a^{2}  b^{3} (FALSE)
This is not an identity. 
4. Determine whether (a + b)^{4} + (a  b)^{4} = 2a^{4} + 12a^{2}b^{2} + 2b^{4} is an identity.
First, expand ( a + b )^{4} 
( a + b )( a + b )( a + b )( a + b)
=
(a^{2} + 2ab + b^{2})(a^{2} + 2ab + b^{2})
= a^{4} + 4a^{3}b + 6a^{2}b^{2} + 4ab^{3} + b^{4}

Now, expand ( a  b )^{4} 
( a  b )( a  b )( a  b )( a  b)
=
(a^{2}  2ab + b^{2})(a^{2}  2ab + b^{2})
= a^{4}  4a^{3}b + 6a^{2}b^{2}  4ab^{3} + b^{4}

Add the results. 
(a^{4} + 4a^{3}b + 6a^{2}b^{2} + 4ab^{3} + b^{4}) added to
(a^{4}  4a^{3}b + 6a^{2}b^{2}  4ab^{3} + b^{4 }) gives
2a^{4}+ 12a^{2}b^{2} + 2b^{4}

This is an identity. 

You may be asked to "check" someone's work to see if they properly showed the existence of an identity.
5. Riley gave the following justification that n^{3}  n = n(n + 1)(n  1) is an identity.
Is this correct? If not, where is the error?

n(n + 1)(n  1) 
Right side of equation. 
Step 1: 
= (n^{2} + n)(n  1) 
Multiply first two terms. 
Step2: 
= n^{3 }+ n^{2}  n^{2} n 
Multiplication (distribution). 
Step 3: 
= n^{3 } n 
Combining like terms. 
n^{3}  n = n(n + 1)(n  1) is an identity. 

Riley's solution is correct!