Abbreviated multiplication formulas or rules are used in arithmetic, more specifically algebra, to speed up the process of evaluating large algebraic expressions. The formulas themselves are derived from rules existing in algebra for multiplying several polynomials.

The use of these formulas provides a fairly quick solution to various mathematical problems, and also helps to simplify expressions. The rules of algebraic transformations allow you to perform some manipulations with expressions, following which you can obtain on the left side of the equality the expression on the right side, or transform the right side of the equality (to obtain the expression on the left side after the equal sign).

It is convenient to know the formulas used for abbreviated multiplication from memory, since they are often used in solving problems and equations. Below are the main formulas included in this list and their names.

Square of the sum

To calculate the square of the sum, you need to find the sum consisting of the square of the first term, twice the product of the first term and the second and the square of the second. In the form of an expression, this rule is written as follows: (a + c)² = a² + 2ac + c².

Squared difference

To calculate the square of the difference, you need to calculate the sum consisting of the square of the first number, twice the product of the first number and the second (taken with the opposite sign) and the square of the second number. In the form of an expression, this rule looks like this: (a - c)² = a² - 2ac + c².

Difference of squares

The formula for the difference of two numbers squared is equal to the product of the sum of these numbers and their difference. In the form of an expression, this rule looks like this: a² - с² = (a + с)·(a - с).

Cube of sum

To calculate the cube of the sum of two terms, you need to calculate the sum consisting of the cube of the first term, triple the product of the square of the first term and the second, triple the product of the first term and the second squared, and the cube of the second term. In the form of an expression, this rule looks like this: (a + c)³ = a³ + 3a²c + 3ac² + c³.

Sum of cubes

According to the formula, it is equal to the product of the sum of these terms and their incomplete squared difference. In the form of an expression, this rule looks like this: a³ + c³ = (a + c)·(a² - ac + c²).

Example. It is necessary to calculate the volume of a figure formed by adding two cubes. Only the sizes of their sides are known.

If the side values ​​are small, then the calculations are simple.

If the lengths of the sides are expressed in cumbersome numbers, then in this case it is easier to use the “Sum of Cubes” formula, which will greatly simplify the calculations.

Difference cube

The expression for the cubic difference sounds like this: as the sum of the third power of the first term, triple the negative product of the square of the first term by the second, triple the product of the first term by the square of the second and the negative cube of the second term. In the form of a mathematical expression, the cube of the difference looks like this: (a - c)³ = a³ - 3a²c + 3ac² - c³.

Difference of cubes

The difference of cubes formula differs from the sum of cubes by only one sign. Thus, the difference of cubes is a formula equal to the product of the difference of these numbers and their incomplete square of the sum. In the form, the difference of cubes looks like this: a 3 - c 3 = (a - c)(a 2 + ac + c 2).

Example. It is necessary to calculate the volume of the figure that will remain after subtracting the volumetric figure from the volume of the blue cube yellow color, which is also a cube. Only the side size of the small and large cube is known.

If the side values ​​are small, then the calculations are quite simple. And if the lengths of the sides are expressed in significant numbers, then it is worth applying the formula entitled “Difference of cubes” (or “Cube of difference”), which will greatly simplify the calculations.

In previous lessons, we looked at two ways to factor a polynomial: taking the common factor out of brackets and the grouping method.

In this lesson we will look at another way to factor a polynomial using abbreviated multiplication formulas.

We recommend that you write each formula at least 12 times. For better memorization, write down all the abbreviated multiplication formulas on a small cheat sheet.

Let's remember what the difference of cubes formula looks like.

The difference of cubes formula is not very easy to remember, so we recommend using a special method to remember it.

It is important to understand that any abbreviated multiplication formula also works in reverse side.

Let's look at an example. It is necessary to factor the difference of cubes.

Please note that “27a 3” is “(3a) 3”, which means that for the difference of cubes formula, instead of “a” we use “3a”.

We use the difference of cubes formula. In place of “a 3” we have “27a 3”, and in place of “b 3”, as in the formula, there is “b 3”.

Applying the difference of cubes in the opposite direction

Let's look at another example. You need to convert the product of polynomials into the difference of cubes using the abbreviated multiplication formula.

Please note that the product of polynomials “(x − 1)(x 2 + x + 1)” resembles the right side of the difference of cubes formula “”, only instead of “a” there is “x”, and in place of “b” there is “1” .

For “(x − 1)(x 2 + x + 1)” we use the difference of cubes formula in the opposite direction.

Let's look at a more complicated example. It is required to simplify the product of polynomials.

If we compare “(y 2 − 1)(y 4 + y 2 + 1)” with the right side of the difference of cubes formula
« a 3 − b 3 = (a − b)(a 2 + ab + b 2)“, then you can understand that in place of “a” from the first bracket there is “y 2”, and in place of “b” there is “1”.

Difference of squares

Let us derive the formula for the difference of squares $a^2-b^2$.

To do this, remember the following rule:

If we add any monomial to the expression and subtract the same monomial, we get the correct identity.

Let's add to our expression and subtract from it the monomial $ab$:

In total, we get:

That is, the difference between the squares of two monomials is equal to the product of their difference and their sum.

Example 1

Present as a product $(4x)^2-y^2$

\[(4x)^2-y^2=((2x))^2-y^2\]

\[((2x))^2-y^2=\left(2x-y\right)(2x+y)\]

Sum of cubes

Let us derive the formula for the sum of cubes $a^3+b^3$.

Let's take the common factors out of brackets:

Let's take $\left(a+b\right)$ out of brackets:

In total, we get:

That is, the sum of the cubes of two monomials is equal to the product of their sum and the incomplete square of their difference.

Example 2

Present as a product $(8x)^3+y^3$

This expression can be rewritten as follows:

\[(8x)^3+y^3=((2x))^3+y^3\]

Using the difference of squares formula, we get:

\[((2x))^3+y^3=\left(2x+y\right)(4x^2-2xy+y^2)\]

Difference of cubes

Let us derive the formula for difference of cubes $a^3-b^3$.

To do this, we will use the same rule as above.

Let's add to our expression and subtract from it the monomials $a^2b\ and\ (ab)^2$:

Let's take the common factors out of brackets:

Let's take $\left(a-b\right)$ out of brackets:

In total, we get:

That is, the difference of the cubes of two monomials is equal to the product of their difference by the incomplete square of their sum.

Example 3

Present as a product $(8x)^3-y^3$

This expression can be rewritten as follows:

\[(8x)^3-y^3=((2x))^3-y^3\]

Using the difference of squares formula, we get:

\[((2x))^3-y^3=\left(2x-y\right)(4x^2+2xy+y^2)\]

Example of problems using formulas for difference of squares and sum and difference of cubes

Example 4

Factorize.

a) $((a+5))^2-9$

c) $-x^3+\frac(1)(27)$

Solution:

a) $((a+5))^2-9$

\[(((a+5))^2-9=(a+5))^2-3^2\]

Applying the difference of squares formula, we get:

\[((a+5))^2-3^2=\left(a+5-3\right)\left(a+5+3\right)=\left(a+2\right)(a +8)\]

Let's write this expression in the form:

Let's apply the formula of cubes:

c) $-x^3+\frac(1)(27)$

Let's write this expression in the form:

\[-x^3+\frac(1)(27)=(\left(\frac(1)(3)\right))^3-x^3\]

Let's apply the formula of cubes:

\[(\left(\frac(1)(3)\right))^3-x^3=\left(\frac(1)(3)-x\right)\left(\frac(1)( 9)+\frac(x)(3)+x^2\right)\]