Standard Identities

In Mathematics, identity is defined as an equation that holds or valid irrespective of the value chosen for its variables. They are used to simplify or rearrange algebraic expressions. The definition states that two identities are replaceable, so we can replace one identity with another identity at any time. For example, the identity (p + q)² = (p² + q² + 2pq), is true for all the alternatives of p and q, whether they are real or complex numbers.

The identities that are obtained by multiplying one binomial with another binomial is known as standard identities. Read the article below to know more about standard identities.

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What are Standard Identities in Maths?

Standard identities are identities that are introduced by multiplying one binomial with another binomial.

Example,

let us consider (p+q)(p +q) or (p +q)²

(p + q)² = (p + q)(p + q)

(p + q)(p + q) or (p + q)²

p² + pq + pq + q²

p² + q² + 2pq  (as pq = qp)

Hence, (p + q)² = p² + q² + 2pq              (1)

Clearly, we can see this is an identity as the expression on LHS is obtained from the RHS. One may verify that for any values of p and q, the values of two sides will always be equal.

Next, we consider, (p - q)(p - q) or (p - q)²

We can say, (p - q)² = (p - q)(p - q)

p(p - q) q(p - q)

p² - pq - pq + q²

p² + q² - 2pq  as (pq = qp)

Hence, (p - q)² = p² + q² - 2pq         (2)

Next, we consider (p + q)(p - q).  We can say, (p + q)(p - q)= p(p - q) + q(p - q)

(p² - pq + pq - q²) = p² - q²  (as pq = qp)

(p + q)(p - q) = p² - q²             (3)

Finally, we consider (p + q) (p + r) We can say, (p + q) (p + r) = p² + pr + pq + qr

or

(p + q) (p + r) = p² + (q + r)p +pq  (4)

The above-given identities 1, 2, 3, 4 are considered as standard identities.

Algebraic Identities

An algebraic identity is an equation that is valid for any value of its variables.

For example, the identity (p + q)² = p² + 2pq + q² is valid for all values of p and q.

As an identity holds valid for all values of its variables, it is possible to replace one side of equality with another side of equality. For example, in the above-given identity, we can easily substitute instances of (p + q)² = p² + q² + 2pq and vice versa.

Using identity in an intelligent way offers shortcuts to many problems by making algebra easier to operate. Below is a list of some standard algebraic identities.

Standard Algebraic Identities Under Binomial Theorem

Following are some of the standard identities in Algebra under binomial theorem.

Identity 1: (p + q)² = p² + 2pq + q²

Identity 2: (p - q)² = p² + q² - 2pq

Identity 3: (p + q)³ = p³ + q³ + 3pq(p + q)

Identity 4: (p - q)³ = p³ - q³ - 3pq(p - q)

Identity 5: (p + q)⁴ = p⁴ + 4p³q + 6p²q² + 4pq³ + q⁴

Identity 6: (p - q)⁴ = p⁴ - 4p³q + 6p²q² - 4pq³ + q⁴

Standard Algebraic Identities Under Factoring

Following standard algebraic identities are factoring formula:

Identity 1: p² - q² = ( p + q)(p - q)

Identity 2: p³ + q³ = (p + q)(p² + q² - pq)

Identity 3: p³ - q³ = ( p - q)(p² + q² + pq)

Identity 4: p⁴ - q⁴ = (p² + q²)(p² - q²)

Standard Algebraic Identities in Three - Variables

Following standard identities in three- variables are obtained by some factoring and manipulation of terms.

Identity 1: (p + q)(p + r)(q + r) = (p + q + r)(pq + pr +qr) - pqr

Identity 2: p² + q² + r² = (p + q + r)² - 2 (pq + pr + qr)

Identity 3: p + q + r - 3pqr = (p + q + r)(p² + q² + r² - pq - pr - qr)

Solved Examples

1. Using identity, (p + q)² = p² + q² + 2pq

Find (2m + 3n)²

Solution:

As we know, (p + q)² = p² + q² + 2pq

Accordingly, (2m + 3n)² = (2m)² + (3n)² + 2(2m)(3n)

Hence, (2m + 3n)² = 4m² + 9n² + 12mn

2. Using identity, (p - q²) = (p² + q² - 2pq)

Find (2m - 3n)²

Solution:

As we know, (p - q²) = (p² + q² - 2pq)

Accordingly, (2m - 3n)² = (2m)² +(3n)² - 2(2m)(3n)

Hence, (2m + 3n)² = 4m² + 9n² - 12mn

3. Using Identity,(p + q)(p - q) = p² - q²

Find  (2m + 3n) (2m - 3n)

Solution:

As we know, (p + q)(p - q) = p² - q²

Accordingly, (2m + 3n)(2m - 3n)  = (2m)² - (3n)²

(2m + 3n)(2m - 3n)  = 4m² - 9n²