Question

How many Identities are there in algebra?Name them.​

Answer 1

$( {a + b})^{2} = {a}^{2} + {b}^{2} + 2ab \\ \\ {(a - b)}^{2} = {a}^{2} + {b}^{2} - 2ab \\ \\ {a}^{2} - {b}^{2} = (a + b)(a - b) \\ \\ {a + b + c}^{2} = {a}^{2} + {b}^{2} + {c}^{2} + 2ab + 2bc + 2ca \\ \\ {a + b}^{3} = {a}^{3} + {b}^{3} + 3ab(a + b) \\ \\ {a - b}^{3} = {a}^{3} - {b}^{3} - 3ab(a - b) \\ \\ {a}^{3} + {b}^{3} + {c}^{3} - 3abc = (a + b + c)( {a}^{2} + {b}^{2} + {c}^{2} - ab - bc - ca).$

Answer 2

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

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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)