Do you have to do **exercises with remarkable identities**? Below you have the formulas that you should know to solve problems with these algebraic expressions that, sometimes, can have a quite laborious development, so knowing their formula will save you a lot of time in their calculation.

Article sections

## Binomial squared

The **binomial squared** is one of the most typical notable identities and yet it is often poorly resolved.

These are the** formulas to be applied** to solve a remarkable identity of a binomial squared:

(a + b)

^{2}= a^{2}+ 2ab + b^{2}(a - b)

^{2}= a^{2}- 2ab + b^{2}

The terms that are squared always have a positive sign while the term that is equal to twice the first times the second will have a sign that will be positive or negative depending on the sign in the remarkable identity.

Solved exercises:

(x + 2)

^{2}= x^{2}+ 4x + 4(x - 2)

^{2}= x^{2}- 4x + 4

## Sum by difference

When we learn the remarkable identity of the sum by difference we are taught that **is equal to the difference of squares**. This is summarized in the following formula:

(a + b) (a - b) = a

^{2}- b^{2}

Solved examples:

(3x + 2) (3x - 2) = 9x

^{2}– 4(2x + 5) (2x - 5) = 4x

^{2}– 25

## Binomial cubed

The **cubed binomial** is one of the most time-consuming remarkable identities to solve, so knowing the formulas is vital to save time.

As is the case with the remarkable identity of the binomial squared, **we will have to use one formula or the other depending on the sign** we have:

(a + b)

^{3}= a^{3}+ 3a^{2}b + 3ab^{2}+ b^{3}(a - b)

^{3}= a^{3}- 3a^{2}b + 3ab^{2}- b^{3}

Below we will look at several **solved examples of how to calculate the remarkable identity** of a cubed binomial:

(x + 3)

^{3}= x^{3}+ 3 - x^{2}- 3 + 3 - x - 3^{2 }+ 3^{3}= x^{3}+ 9x^{2}+ 27x + 27(2x - 3)

^{3}= (2x)^{3}- 3 - (2x)^{2}- 3 + 3 - 2x - 3^{2}− 3^{3}= 8x^{3}- 36x^{2}+ 54x - 27

## Trinomial squared

In this case **we have the sum of three variables, the result of which we square**a remarkable identity that we can calculate by applying the following formula:

(a + b + c)

^{2}= a^{2}+ b^{2}+ c^{2 }+ 2ab + 2ac + 2bc

Again, let's see with a solved example how this remarkable identity is made:

(x

^{2}+ x + 2) = x^{4}+ x^{2}+ 4 + 2x^{3}+ 4x^{2}+ 4x = x^{4}+ 2x^{3}+ 5x^{2}+ 4x + 4

## Sum of cubes

If we have the **sum of two variables cubed**The way to calculate this remarkable identity is to apply the following formula:

a

^{3}+ b^{3}= (a + b) - (a^{2}- ab + b^{2})

Let's see it with a practical example:

64x

^{3}+ 27 = (4x + 3) (16x^{2}- 12x + 9)

## Difference of cubes

In the case that instead of the sum of two cubed variables** let's have a subtraction**the formula to be applied is as follows:

a

^{3}- b^{3}= (a - b) - (a^{2}+ ab + b^{2})

If we change the sign of the previous example to make it compatible with the difference of cubes, we have that:

64x

^{3}- 27 = (4x - 3) (16x^{2}+ 12x + 9)

## Product of two binomials with a common term

We finish the section dedicated to remarkable identities with a** product of two binomials having a common term** (the x). In this case, we have to apply the following formula to solve the equation:

(x + a) (x + b) = x

^{2}+ (a + b) x + ab

As it could not be otherwise, let's see it better with a practical example:

(x + 1) (x + 2) = x

^{2}+ 3x + 2

If you have any doubts about **exercise of remarkable identities**Leave us a comment and we will try to help you as soon as possible.

In this case, we have not been able to develop a **remarkable identities calculator** as it is too complex due to the large number of variations involved in working with formulas involving numbers and variables whose value is unknown (x). Therefore, we will be happy to help you if you have any doubts with a particular remarkable identity.