# Simplify and write (x2 – y2)(x-2 + y-2) using positive exponents

Last Updated : 25 Dec, 2023

Algebra is the branch of mathematics that deals with the study of various symbols which represent such quantities as do not have a constant value or quantity associated with them, instead they tend to vary or change over time with respect to some other factor. Such symbols are regarded as variables in the study of algebra, and the quantities attached to them are called coefficients. They can be depicted through various shapes or even English alphabets. In other words, algebra considers representing numbers through letters or symbols without laying emphasis on depicting their actual values.

### Algebraic Expression

An algebraic expression is such a statement that is formed using variables and constants in mathematics, along with various arithmetic operations like addition, subtraction, multiplication, division, exponential operation, root extraction like square root, cube root, fourth root, so on and so forth.

Examples:

• x + 1 is an algebraic expression with x as the variable and addition as the operation.
• x2 âˆ’ 1 is an algebraic expression with x as the variable and subtraction and exponent as the operation.
• 2x2 âˆ’ 3xy + 5 is an algebraic expression with x and y as the variables with addition, exponent, subtraction and multiplication as the operations.

Basic Terminology

• Variable: A variable is such a term in an algebraic expression that can assume any value, its real value does not exist.
• Coefficient: It is a constant and well- defined which is always used with the variable.
• Operator: It means any arithmetic operation like addition, subtraction, multiplication, division, exponential operation, root extraction like square root, cube root, fourth root, so on and so forth.
• Constant: Such a term that is independent of both coefficient and the variable and is well-defined in itself is called the constant.
• Exponent: The number of times a number has been multiplied by itself refers to its exponent.

### Rules of Exponents

Rule 1: If two or more bases have the same powers and are in multiplication, their powers are added together keeping the base intact, i.e., am Ã— an = am+n.

Example:

• 23 Ã— 25 = 23+5 = 28
• 4-2 Ã— 43 Ã— 4100 = 4-2+3+100 = 4101

Rule 2: If two or more bases have the same powers and are in the division, their powers are subtracted together keeping the base intact. It is to be noted that the power of the denominator is to be deducted from that power of the numerator, i.e., am Ã· an = am-n.

Example:

•  = 24-3 = 21 = 2
•  = 104-8 = 10-4

Rule 3: Anything raised to the power zero equals 1.

Example:

• 20 = 1
• 10000000 = 1
• 8590 = 1

Rule 4: When the power of an exponent already raised to a power is given, one needs to multiply those powers together, i.e., (am)n = amn.

Example:

• (23)4 = (2)3Ã—4 = 212
• [(-3)-9]Â² = (-3)-9Ã—2 = (-3)-18

Rule 5: Where two different bases have the same power, the bases are multiplied and the product is raised to the power both the bases had before multiplication, i.e., am Ã— bm = (a Ã— b)m.

Example:

• 43 Ã— 103 = (4 Ã— 10)3 = 403
• 2123 Ã— 56123 = (2 Ã— 56)123 = 112123

Rule 6: In case we are given a fractional exponent, then the numerator becomes the power of the base and the denominator is taken as the root of the whole expression, i.e., am/n

Example:

• 21/2
• 21/3
• 24/5

Rule 7: If the power is negative, reciprocate the base to make it positive, i.e., a-m.

Example:

• 2-9
• 100-8

### Simplify and write (x<sup>2</sup> – y<sup>2</sup>)(x<sup>-2</sup> + y<sup>-2</sup>) using positive exponents

x2(x-2) – y2(x-2) + x2(y-2) -y2(y-2)

Using the properties am . an = am+n and a-x, we have

= x2+(-2) –  – y2+(-2)

= x0 – y0

Anything raised to the power 0 is always 1, so

–  + 1 -1

=

### Similar Problems

Question 1. Simplify: (x-7 y10) (x-8 y3).

Solution:

= (x-7 x-8) (y10 y3)

Using the property am . an = am+n

= x-7-8 y10+3

= x-15 y13

Using a-m = 1/ am, we have:

= y13/ x15

Question 2. Simplify: .

Solution:

= 3/5 (x2/x3) (y3/y2)

= 3/5 (x2-3) (y3-2)

= 3/5 (x-1) (y1)

Using the property a-m = 1/ am, we have:

= 3y/5x

Question 3. Simplify and write as positive exponents: (64x-6 y6)5/6.

Solution:

Using the property (abc)m = am bm cm, we have:

(64x-6 y6)5/6 = 645/6 x-30/6 y30/6

= 25 x-6 y5

Using the property a-m = 1/am, we have:

Hence, (64x-6 y6)5/6 = 32y5/ x6

Question 4. Simplify and write as positive exponents: .

Solution:

Since anything raised to the power zero is always 1.

Thus, 4y0 = 1

Hence the expression becomes 2x-10/ 1.

= 2x-10

Since a-m = 1/am,

Hence,  = 2/ x10.

Question 5. Simplify: .

Solution:

Using the property (am)n = amn, we have:

[(2a3)/ (3a5 Ã— 3]3

= [(2a3)/ (3a15]3

Using the property am/an = am-n, we have:

= 2/3[a3-15]3

= [2[a-12]3]/3

= 2a-36/3

Hence, [(2a3)/ (3a5)3]3 = .