Formula for reducing a cube. Abbreviated multiplication formulas. Conclusions from the lesson

Abbreviated multiplication formulas (FMF) are used to exponentiate and multiply numbers and expressions. Often these formulas allow you to make calculations more compactly and quickly.

In this article we will list the basic formulas for abbreviated multiplication, group them in a table, consider examples of using these formulas, and also dwell on the principles of proof of formulas for abbreviated multiplication.

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For the first time, the topic of FSU is considered within the framework of the Algebra course for the 7th grade. Below are 7 basic formulas.

Abbreviated multiplication formulas

1. formula for the square of the sum: a + b 2 = a 2 + 2 a b + b 2
2. square difference formula: a - b 2 = a 2 - 2 a b + b 2
3. sum cube formula: a + b 3 = a 3 + 3 a 2 b + 3 a b 2 + b 3
4. difference cube formula: a - b 3 = a 3 - 3 a 2 b + 3 a b 2 - b 3
5. square difference formula: a 2 - b 2 = a - b a + b
6. formula for sum of cubes: a 3 + b 3 = a + b a 2 - a b + b 2
7. formula for difference of cubes: a 3 - b 3 = a - b a 2 + a b + b 2

The letters a, b, c in these expressions can be any numbers, variables or expressions. For ease of use, it is better to learn the seven basic formulas by heart. Let's put them in a table and present them below, encircling them with a frame.

The first four formulas allow you to calculate, respectively, the square or cube of the sum or difference of two expressions.

The fifth formula calculates the difference between the squares of expressions by multiplying their sum and difference.

The sixth and seventh formulas are, respectively, multiplying the sum and difference of expressions by the incomplete square of the difference and the incomplete square of the sum.

The abbreviated multiplication formula is sometimes also called the abbreviated multiplication identities. This is not surprising, since every equality is an identity.

When solving practical examples, abbreviated multiplication formulas with the left and right sides swapped are often used. This is especially convenient when factoring a polynomial.

Additional abbreviated multiplication formulas

Let's not limit ourselves to the 7th grade algebra course and add a few more formulas to our FSU table.

First, let's look at Newton's binomial formula.

a + b n = C n 0 · a n + C n 1 · a n - 1 · b + C n 2 · a n - 2 · b 2 + . . + C n n - 1 · a · b n - 1 + C n n · b n

Here C n k are the binomial coefficients that appear in line number n in Pascal’s triangle. Binomial coefficients are calculated using the formula:

C n k = n ! k! · (n - k) ! = n (n - 1) (n - 2) . . (n - (k - 1)) k !

As we can see, the FSF for the square and cube of the difference and the sum is a special case of the Newton binomial formula for n=2 and n=3, respectively.

But what if there are more than two terms in the sum that needs to be raised to a power? The formula for the square of the sum of three, four or more terms will be useful.

a 1 + a 2 + . . + a n 2 = a 1 2 + a 2 2 + . . + a n 2 + 2 a 1 a 2 + 2 a 1 a 3 + . . + 2 a 1 a n + 2 a 2 a 3 + 2 a 2 a 4 + . . + 2 a 2 a n + 2 a n - 1 a n

Another formula that may be useful is the formula for the difference between the nth powers of two terms.

a n - b n = a - b a n - 1 + a n - 2 b + a n - 3 b 2 + . . + a 2 b n - 2 + b n - 1

This formula is usually divided into two formulas - for even and odd powers, respectively.

For even 2m indicators:

a 2 m - b 2 m = a 2 - b 2 a 2 m - 2 + a 2 m - 4 b 2 + a 2 m - 6 b 4 + . . + b 2 m - 2

For odd exponents 2m+1:

a 2 m + 1 - b 2 m + 1 = a 2 - b 2 a 2 m + a 2 m - 1 b + a 2 m - 2 b 2 + . . + b 2 m

The difference of squares and difference of cubes formulas, as you guessed, are special cases of this formula for n = 2 and n = 3, respectively. For difference of cubes, b is also replaced by - b.

How to read abbreviated multiplication formulas?

We will give the appropriate formulations for each formula, but first we will understand the principle of reading formulas. The most convenient way to do this is with an example. Let's take the very first formula for the square of the sum of two numbers.

a + b 2 = a 2 + 2 a b + b 2 .

They say: the square of the sum of two expressions a and b is equal to the sum of the square of the first expression, twice the product of the expressions and the square of the second expression.

All other formulas are read similarly. For the square of the difference a - b 2 = a 2 - 2 a b + b 2 we write:

the square of the difference between two expressions a and b is equal to the sum of the squares of these expressions minus twice the product of the first and second expressions.

Let's read the formula a + b 3 = a 3 + 3 a 2 b + 3 a b 2 + b 3. The cube of the sum of two expressions a and b is equal to the sum of the cubes of these expressions, triple the product of the square of the first expression by the second, and triple the product of the square of the second expression by the first expression.

Let's move on to reading the formula for the difference of cubes a - b 3 = a 3 - 3 a 2 b + 3 a b 2 - b 3. The cube of the difference between two expressions a and b is equal to the cube of the first expression minus the triple product of the square of the first expression and the second, plus the triple product of the square of the second expression and the first expression, minus the cube of the second expression.

The fifth formula a 2 - b 2 = a - b a + b (difference of squares) reads like this: the difference of the squares of two expressions is equal to the product of the difference and the sum of the two expressions.

For convenience, expressions like a 2 + a b + b 2 and a 2 - a b + b 2 are called, respectively, the incomplete square of the sum and the incomplete square of the difference.

Taking this into account, the formulas for the sum and difference of cubes can be read as follows:

The sum of the cubes of two expressions is equal to the product of the sum of these expressions and the partial square of their difference.

The difference between the cubes of two expressions is equal to the product of the difference between these expressions and the partial square of their sum.

Proof of the FSU

Proving FSU is quite simple. Based on the properties of multiplication, we will multiply the parts of the formulas in brackets.

For example, consider the formula for the squared difference.

a - b 2 = a 2 - 2 a b + b 2 .

To raise an expression to the second power, you need to multiply this expression by itself.

a - b 2 = a - b a - b .

Let's expand the brackets:

a - b a - b = a 2 - a b - b a + b 2 = a 2 - 2 a b + b 2 .

The formula is proven. The remaining FSUs are proven similarly.

Examples of FSU application

The purpose of using abbreviated multiplication formulas is to quickly and concisely multiply and raise expressions to powers. However, this is not the entire scope of application of the FSU. They are widely used in reducing expressions, reducing fractions, and factoring polynomials. Let's give examples.

Example 1. FSU

Let's simplify the expression 9 y - (1 + 3 y) 2.

Let's apply the sum of squares formula and get:

9 y - (1 + 3 y) 2 = 9 y - (1 + 6 y + 9 y 2) = 9 y - 1 - 6 y - 9 y 2 = 3 y - 1 - 9 y 2

Example 2. FSU

Let's reduce the fraction 8 x 3 - z 6 4 x 2 - z 4.

We note that the expression in the numerator is the difference of cubes, and in the denominator is the difference of squares.

8 x 3 - z 6 4 x 2 - z 4 = 2 x - z (4 x 2 + 2 x z + z 4) 2 x - z 2 x + z .

We reduce and get:

8 x 3 - z 6 4 x 2 - z 4 = (4 x 2 + 2 x z + z 4) 2 x + z

FSUs also help calculate the values ​​of expressions. The main thing is to be able to notice where to apply the formula. Let's show this with an example.

Let's square the number 79. Instead of cumbersome calculations, let's write:

79 = 80 - 1 ; 79 2 = 80 - 1 2 = 6400 - 160 + 1 = 6241 .

It would seem that a complex calculation is carried out quickly just using abbreviated multiplication formulas and a multiplication table.

Another important point is the selection of the square of the binomial. The expression 4 x 2 + 4 x - 3 can be converted into 2 x 2 + 2 · 2 · x · 1 + 1 2 - 4 = 2 x + 1 2 - 4 . Such transformations are widely used in integration.

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In order to simplify algebraic polynomials, there are abbreviated multiplication formulas. There are not so many of them and they are easy to remember, but you need to remember them. The notation used in formulas can take any form (number or polynomial).

The first abbreviated multiplication formula is called difference of squares. It consists in subtracting the square of one number from the square of the second number, which is equal to the difference between these numbers, as well as their product.

a 2 - b 2 = (a - b)(a + b)

Let's look at it for clarity:

22 2 - 4 2 = (22-4)(22+4)=18 * 26 = 468
9a 2 - 4b 2 c 2 = (3a - 2bc)(3a + 2bc)

The second formula is about sum of squares. It sounds like the sum of two quantities squared is equal to the square of the first quantity, the double product of the first quantity multiplied by the second is added to it, the square of the second quantity is added to them.

(a + b) 2 = a 2 +2ab + b 2

Thanks to this formula, it becomes much easier to calculate the square of a large number, without the use of computer technology.

So for example: the square of 112 will be equal to
1) First, let’s break down 112 into numbers whose squares are familiar to us
112 = 100 + 12
2) We enter the result in squared brackets
112 2 = (100+12) 2
3) Applying the formula, we get:
112 2 = (100+12) 2 = 100 2 + 2 * 100 * 12 + 122 = 10000 + 2400+ 144 = 12544

The third formula is squared difference. Which says that two quantities subtracted from each other in a square are equal, because from the first quantity squared we subtract the double product of the first quantity multiplied by the second, adding to them the square of the second quantity.

(a + b) 2 = a 2 - 2ab + b 2

where (a - b) 2 equals (b - a) 2. To prove this, (a-b) 2 = a 2 -2ab+b 2 = b 2 -2ab + a 2 = (b-a) 2

The fourth formula for abbreviated multiplication is called cube of sum. Which sounds like: two summand quantities in a cube are equal to the cube of 1 quantity, the triple product of 1 quantity squared multiplied by the 2nd quantity is added, to these is added the triple product of 1 quantity multiplied by the square of 2 quantities, plus the second quantity cubed.

(a+b) 3 = a 3 + 3a 2 b + 3ab 2 + b 3

The fifth, as you already understood, is called difference cube. Which finds the differences between quantities, as from the first notation in the cube we subtract the triple product of the first notation in the square multiplied by the second, to them is added the triple product of the first notation multiplied by the square of the second notation, minus the second notation in the cube.

(a-b) 3 = a 3 - 3a 2 b + 3ab 2 - b 3

The sixth is called - sum of cubes. The sum of the cubes is equal to the product of the two addends multiplied by the partial square of the difference, since there is no double value in the middle.

a 3 + b 3 = (a+b)(a 2 -ab+b 2)

Another way to say the sum of cubes is to call it the product in two brackets.

The seventh and final one is called difference of cubes(it can be easily confused with the difference cube formula, but these are different things). The difference of cubes is equal to the product of the difference of two quantities multiplied by the partial square of the sum, since there is no double value in the middle.

a 3 - b 3 = (a-b)(a 2 +ab+b 2)

And so there are only 7 formulas for abbreviated multiplication, they are similar to each other and are easy to remember, the only important thing is not to get confused in the signs. They are also designed to be used in reverse order, and textbooks contain quite a few such tasks. Be careful and everything will work out for you.

If you have questions about the formulas, be sure to write them in the comments. We will be happy to answer you!

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In the previous lesson we dealt with factorization. We mastered two methods: putting the common factor out of brackets and grouping. In this lesson - the following powerful method: abbreviated multiplication formulas. In short - FSU.

Abbreviated multiplication formulas (sum and difference square, sum and difference cube, difference of squares, sum and difference of cubes) are extremely necessary in all branches of mathematics. They are used in simplifying expressions, solving equations, multiplying polynomials, reducing fractions, solving integrals, etc. and so on. In short, there is every reason to deal with them. Understand where they come from, why they are needed, how to remember them and how to apply them.

Do we understand?)

Where do abbreviated multiplication formulas come from?

Equalities 6 and 7 are not written in a very familiar way. It's kind of the opposite. This is on purpose.) Any equality works both from left to right and from right to left. This entry makes it clearer where the FSUs come from.

They are taken from multiplication.) For example:

(a+b) 2 =(a+b)(a+b)=a 2 +ab+ba+b 2 =a 2 +2ab+b 2

That's it, no scientific tricks. We simply multiply the brackets and give similar ones. This is how it turns out all abbreviated multiplication formulas. Abbreviated multiplication is because in the formulas themselves there is no multiplication of brackets and reduction of similar ones. Abbreviated.) The result is immediately given.

FSU needs to be known by heart. Without the first three, you can’t dream of a C; without the rest, you can’t dream of a B or A.)

Why do we need abbreviated multiplication formulas?

There are two reasons to learn, even memorize, these formulas. The first is that a ready-made answer automatically reduces the number of errors. But this is not the main reason. But the second one...

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You can get acquainted with functions and derivatives.

One of the first topics studied in an algebra course is abbreviated multiplication formulas. In grade 7, they are used in the simplest situations, where you need to recognize one of the formulas in an expression and factor a polynomial or, conversely, quickly square or cube a sum or difference. In the future, FSU is used to quickly solve inequalities and equations and even to calculate some numerical expressions without a calculator.

What does a list of formulas look like?

There are 7 basic formulas that allow you to quickly multiply polynomials in brackets.

Sometimes this list also includes an expansion for the fourth degree, which follows from the presented identities and has the form:

a⁴ — b⁴ = (a - b)(a + b)(a² + b²).

All equalities have a pair (sum - difference), except the difference of squares. The formula for the sum of squares is not given.

The remaining equalities are easy to remember:

It should be remembered that FSUs work in any case and for any values a And b: these can be either arbitrary numbers or integer expressions.

In a situation where you suddenly can’t remember which sign is in front of a particular term in the formula, you can open the brackets and get the same result as after using the formula. For example, if a problem arose when applying the difference cube FSU, you need to write down the original expression and perform multiplication one by one:

(a - b)³ = (a - b)(a - b)(a - b) = (a² - ab - ab + b²)(a - b) = a³ - a²b - a²b + ab² - a²b + ab² + ab² - b³ = a³ - 3a²b + 3ab² - b³.

As a result, after bringing all similar terms, the same polynomial as in the table was obtained. The same manipulations can be carried out with all other FSUs.

Application of FSU to solve equations

For example, you need to solve an equation containing polynomial of degree 3:

x³ + 3x² + 3x + 1 = 0.

The school curriculum does not cover universal techniques for solving cubic equations, and such tasks are most often solved using simpler methods (for example, factorization). If we notice that the left side of the identity resembles the cube of a sum, then the equation can be written in a simpler form:

(x + 1)³ = 0.

The root of such an equation is calculated orally: x = -1.

Inequalities are solved in a similar way. For example, you can solve the inequality x³ – 6x² + 9x > 0.

First of all, you need to factor the expression. First you need to bracket x. After this, note that the expression in parentheses can be converted to the square of the difference.

Then you need to find the points at which the expression takes zero values ​​and mark them on the number line. In a particular case, these will be 0 and 3. Then, using the interval method, determine in which intervals x will correspond to the inequality condition.

FSUs may be useful when performing some calculations without the help of a calculator:

703² - 203² = (703 + 203)(703 - 203) = 906 ∙ 500 = 453000.

Additionally, by factoring expressions, you can easily reduce fractions and simplify various algebraic expressions.

Examples of problems for grades 7-8

In conclusion, we will analyze and solve two tasks on the use of abbreviated multiplication formulas in algebra.

Task 1. Simplify the expression:

(m + 3)² + (3m + 1)(3m - 1) - 2m (5m + 3).

Solution. The condition of the task requires simplifying the expression, i.e. opening the parentheses, performing the operations of multiplication and exponentiation, and also bringing all similar terms. Let us conditionally divide the expression into three parts (according to the number of terms) and open the brackets one by one, using FSU where possible.

• (m + 3)² = m² + 6m + 9(sum square);
• (3m + 1)(3m - 1) = 9m² – 1(difference of squares);
• In the last term you need to multiply: 2m (5m + 3) = 10m² + 6m.

Let's substitute the results obtained into the original expression:

(m² + 6m + 9) + (9m² – 1) - (10m² + 6m).

Taking into account the signs, we will open the brackets and present similar terms:

m² + 6m + 9 + 9m² 1 - 10m² – 6m = 8.

Problem 2. Solve an equation containing the unknown k to the 5th power:

k⁵ + 4k⁴ + 4k³ – 4k² – 4k = k³.

Solution. In this case, it is necessary to use the FSU and the grouping method. It is necessary to move the last and penultimate terms to the right side of the identity.

k⁵ + 4k⁴ + 4k³ = k³ + 4k² + 4k.

The common factor is derived from the right and left sides (k² + 4k +4):

k³(k² + 4k + 4) = k (k² + 4k + 4).

Everything is transferred to the left side of the equation so that 0 remains on the right:

k³(k² + 4k + 4) - k (k² + 4k + 4) = 0.

Again it is necessary to take out the common factor:

(k³ - k)(k² + 4k + 4) = 0.

From the first factor obtained we can derive k. According to the short multiplication formula, the second factor will be identically equal to (k+2)²:

k (k² - 1)(k + 2)² = 0.

Using the difference of squares formula:

k (k - 1)(k + 1)(k + 2)² = 0.

Since a product is equal to 0 if at least one of its factors is zero, finding all the roots of the equation is not difficult:

1. k = 0;
2. k - 1 = 0; k = 1;
3. k + 1 = 0; k = -1;
4. (k + 2)² = 0; k = -2.

Based on illustrative examples, you can understand how to remember formulas, their differences, and also solve several practical problems using FSU. The tasks are simple and there should be no difficulties in completing them.