Answer:
the technical number depends on what you are including to be algebraic identities.
There are multiple forms of most identities (pos. / neg. and multiple ways of writing the same identity)
generally, you can consider there to be 10
Step-by-step explanation:
algebraic identity: an equality that holds true for any variable values
standard identities:
square of binomial:
(a+b)² = a² +2ab + b²
(a - b)² = a² - 2ab + b²
difference of squares:
(a + b)(a - b) = a² - b²
----
product of two binomials:
(x + a)(x + b) = x² + (a + b)x + ab
square of trinomial:
(x + y + z)² = x² + y² + z² + 2xy + 2yz + 2zx
(x - y - z)² = x² + y² + z² -2(xy + yz - zx)
cube of binomial:
(x + y)³ = x³ + y³ + 3xy (x + y)
(x - y)³ = x³ + y³ - 3xy (x - y)
(a - b)³ = a³ - 3a²b + 3ab² - b³
sum of cubes:
a³ + b³ = (a + b)(a² - ab + b²)
x³ + y³ = (x + y)(x² - xy + y²)
x³ + y³ = (x - y)(x² + xy + y²)
difference of cubes:
a³ - b³ = (a - b)(a² + ab + b²)
x³ - y³ = (x - y)³ + 3xy(x - y)
--
x³ + y³ + z³ - 3xyz = (x + y + z)(x² + y² + z² - xy - yz - zx)
= 1/2 (x + y + z) [(x -y)² + (y - z)² + (z - x)²]
(a + b + c)³ = a³ + b³ + c³ + 3(a + b)(b + c)(c + a)
--
and
if x + y + z = 0, then x³ + y³ + z³ = 3xyz
hope this helps!!
(this took me a while to write; there are also a few complicated identities also that I've left out)
