Formulas or rules of abbreviated multiplication are used in arithmetic, or rather in algebra, for a faster process of calculating large algebraic expressions. The formulas themselves are derived from the rules existing in algebra for multiplying several polynomials.

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

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

Sum squared

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 by the second, and the square of the second. As an expression, this rule is written as follows: (a + c) ² \u003d a² + 2ac + c².

Difference squared

To calculate the square of the difference, it is necessary to calculate the sum consisting of the square of the first number, twice the product of the first number by the second (taken with the opposite sign) and the square of the second number. As an expression, this rule looks like this: (a - c) ² \u003d a² - 2ac + c².

Difference of squares

The formula for the difference between two numbers squared is equal to the product of the sum of these numbers by their difference. As an expression, this rule looks as follows: a² - c² \u003d (a + c) · (a - c).

Sum cube

To calculate the cube of the sum of two terms, it is necessary to calculate the sum consisting of the cube of the first term, triple product of the square of the first term and the second, triple product of the first term and the second squared, as well as the cube of the second term. As an expression, this rule looks like this: (a + c) ³ \u003d a³ + 3a²c + 3ac² + c³.

Sum of cubes

According to the formula, it is equated to the product of the sum of these terms by their incomplete square of the difference. In the form of an expression, this rule looks as follows: a³ + c³ \u003d (a + c) · (a² - ac + c²).

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

If the side values \u200b\u200bare small, then the calculations are easy.

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

Difference cube

The expression for the cubic difference is as follows: 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 as follows: (a - c) ³ \u003d a³ - 3a²c + 3ac² - c³.

Difference cubes

The formula for the difference of cubes differs from the sum of cubes in only one sign. Thus, the difference between the cubes is a formula equal to the product of the difference of these numbers by their incomplete square of the sum. In the form, the difference of cubes is as follows: a 3 - c 3 \u003d (a - c) (a 2 + ac + c 2).

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

If the side values \u200b\u200bare small, the calculations are fairly straightforward. And if the lengths of the sides are expressed in significant numbers, then it is worth using a formula entitled "Difference Cubes" (or "Difference Cube"), which will greatly simplify the calculations.

In the previous lessons, we looked at two ways to factor a polynomial into factors: parentheses and grouping.

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

We recommend prescribing each formula at least 12 times. For better memorization, write out all the formulas for abbreviated multiplication for yourself on a small cheat sheet.

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

The formula for the difference between cubes is not very easy to memorize, so we recommend using a special way to memorize it.

It is important to understand that any formula for abbreviated multiplication works in back side.

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

Note that "27a 3" is "(3a) 3", which means that for the formula for the difference between cubes, instead of "a" we use "3a".

We use the formula for the difference of cubes. In place "a 3" we have "27a 3", and in place "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 want to convert the product of polynomials to 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 formula for the difference between cubes "", only instead of "a" there is "x", and instead of "b" there is "1" ...

We use for "(x - 1) (x 2 + x + 1)" the formula for the difference of cubes 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 cubes difference formula
« a 3 - b 3 \u003d (a - b) (a 2 + ab + b 2)", Then you can understand that in the place" a "from the first bracket is" y 2, and in place "b" is "1".

Difference of squares

We 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, then we get the correct identity.

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

Total, we get:

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

Example 1

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

\\ [(4x) ^ 2-y ^ 2 \u003d ((2x)) ^ 2-y ^ 2 \\]

\\ [((2x)) ^ 2-y ^ 2 \u003d \\ left (2x-y \\ right) (2x + y) \\]

Sum of cubes

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

Factor out the common factors:

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

Total, we get:

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

Example 2

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

This expression can be rewritten as follows:

\\ [(8x) ^ 3 + y ^ 3 \u003d ((2x)) ^ 3 + y ^ 3 \\]

Using the formula for the difference of squares, we get:

\\ [((2x)) ^ 3 + y ^ 3 \u003d \\ left (2x + y \\ right) (4x ^ 2-2xy + y ^ 2) \\]

Difference cubes

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

For this, we will use the same rule as above.

Add to our expression and subtract from it the monomials $ a ^ 2b \\ and \\ (ab) ^ 2 $:

Factor out the common factors:

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

Total, we get:

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

Example 3

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

This expression can be rewritten as follows:

\\ [(8x) ^ 3-y ^ 3 \u003d ((2x)) ^ 3-y ^ 3 \\]

Using the formula for the difference of squares, we get:

\\ [((2x)) ^ 3-y ^ 3 \u003d \\ left (2x-y \\ right) (4x ^ 2 + 2xy + y ^ 2) \\]

An example of problems using the formulas for the difference of squares and the sum and difference of cubes

Example 4

Factor out.

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

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

Decision:

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

\\ [(((a + 5)) ^ 2-9 \u003d (a + 5)) ^ 2-3 ^ 2 \\]

Applying the formula for the difference of squares, we get:

\\ [((a + 5)) ^ 2-3 ^ 2 \u003d \\ left (a + 5-3 \\ right) \\ left (a + 5 + 3 \\ right) \u003d \\ left (a + 2 \\ right) (a +8) \\]

Let's write this expression in the form:

Let's apply the kuma cubes formula:

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

Let's write this expression in the form:

\\ [- x ^ 3 + \\ frac (1) (27) \u003d (\\ left (\\ frac (1) (3) \\ right)) ^ 3-x ^ 3 \\]

Let's apply the kuma cubes formula:

\\ [(\\ left (\\ frac (1) (3) \\ right)) ^ 3-x ^ 3 \u003d \\ left (\\ frac (1) (3) -x \\ right) \\ left (\\ frac (1) ( 9) + \\ frac (x) (3) + x ^ 2 \\ right) \\]