How to prove algebraic identities: A comprehensive Guide
- Author: Noreen Niazi
- Last Updated on: February 1, 2024
Algebraic identities are the backbone of mathematics. You need to learn about the algebraic identities. But how to prove algebraic identities? Let’s discuss this in detail.
If you are a student or teacher and looking for an easy way to prove algebraic identities, then this blog post is for you. Let’s go through this blog post and reveal an easy way to prove the algebraic identities.
What are algebraic identities?
Algebraic identities are the equations that are true for all values of variables.
For example, consider the $$a^2-b^2=(a+b)(a-b)$$. Now here you can take any value, result is true for all values.
Let $$a=2$$ and $$b=3$$ then
$$a^2-b^2=(a+b)(a-b)$$
$$a=2$$ and $$b=3$$. Let’s first solve the left-hand side
$$L.H.S=a^2-b^2$$
$$=2^2-3^2$$
$$=4-9$$
$$a^2-b^2=-5$$
Now solve the right-hand side.
$$R.H.S=(a+b)(a-b)$$
$$a=2, b=3$$
$$=(2+3)(2-3)$$
$$=(5)(-1)$$
$$(a+b)(a-b)=-5$$
Here left-hand side is equal to the right-hand side.
$$a^2-b^2=(a+b)(a-b)$$
How to prove algebraic identities Algebraically:
The whole process of proving the identity is about realizing that LHS and RHS are equal for all possible values of variables. Here’s a step-by-step guide:
- Identify the Identity: The first step is to determine what you want to prove as your identity. It may be a common identity or even something more complicated.
- Expand the Expressions: Enlarge the RHS and LHS of the equation. The expressions can be simplified using the distributive property $$(a(b + c) = ab+ac)$$, associative property $$( a+( b+c)= ((a)+b))$$ and commutativity rule.
- Simplify the Expressions: Combine the like terms on each side. Like terms are coefficients that share the same variables and powers.
- Equating the Sides: If, after being simplified, the LHS is equal to the RHS then an identity has been proven.
Examples of proving algebraic identities algebraically
Proving algebraic identities becomes a matter of routine with practice. It’s like putting together a puzzle – every step, you get closer to the solution. Therefore, get your pen and paper ready for some proving! Recall that every identity is an uncharted path in the magical land of algebra.
1
Identity 1: $$(x+y)^2=x^2+y^2+2xy$$
The first famous and standard algebraic identity is
$$(x+y)^2=x^2+y^2+2xy$$
Now to prove the identity we solve the left-hand side before and then get the right-hand side:
- $$(x+y)^2=x^2+y^2$$
- Now we expand the expression to get the result.
$$L.H.S=(x+y)^2$$
$$ =(x+y)(x+y)$$
Now by using the distributive we get
$$ =x(x+y)+y(x+y)$$
Again using the distributive property we get
$$ =x^2+xy+yx+y^2$$
Since by commutative property $$xy$$=$$yx$$. Therefore the given expression become
$$ =x^2+2xy+y^2=R.H.S$$
2
Identity 2: $$(x-y)^2=x^2+y^2-2xy$$
The second famous and standard algebraic identity is
$$(x-y)^2=x^2+y^2-2xy$$
Now to prove the identity we solve the left-hand side before and then get the right-hand side:
- $$(x-y)^2=x^2+y^2-2xy$$
- Now we expand the expression to get the result.
$$L.H.S=(x+y)^2$$
$$ =(x-y)(x-y)$$
Now by using the distributive we get
$$ =x(x-y)-y(x-y)$$
Again using the distributive property we get
$$ =x^2-xy-yx+y^2$$
Since by commutative property $$xy$$=$$yx$$. Therefore the given expression become
$$ =x^2-2xy+y^2=R.H.S$$
3
Identity 3: $$(x+y)(x-y)=x^2-y^2$$
The third famous and standard algebraic identity is
$$(x+y)(x-y)=x^2-y^2$$
Now to prove the identity we solve the left-hand side before and then get the right-hand side:
- $$(x+y)(x-y)=x^2-y^2$$
- Now we expand the expression to get the result.
$$L.H.S=(x+y)(x-y)$$
Now by using the distributive we get
$$ =x(x-y)+y(x-y)$$
Again using the distributive property we get
$$ =x^2-xy+yx-y^2$$
Since by commutative property $$xy$$=$$yx$$. Therefore the given expression become
$$ =x^2-y^2=R.H.S$$
4
Identity 4: $$(x+a)(x+b)=x^2+(a+b)x+ab$$
Now to prove the identity we solve the left-hand side before and then get the right-hand side:
- $$(x+a)(x+b)=x^2+(a+b)x+ab$$
- Now we expand the expression to get the result.
$$L.H.S=(x+a)(x+b)$$
Now by using the distributive we get
$$ =x(x+b)+a(x+b)$$
Again using the distributive property we get
$$ =x^2+xb+ax+b^2$$
By taking $$x$$ common the given expression becomes
$$ (x+a)(x+b)=x^2+(a+b)x+ab$$
5
Identity 5: $$(x+y)^2-(x-y)^2=4xy$$
The fifth famous and standard algebraic identity is
$$(x+y)^2-(x-y)^2=4xy$$
Now to prove the identity we solve the left-hand side before and then get the right-hand side:
- $$(x+y)^2-(x-y)^2=4xy$$
- Now we expand the expression to get the result.
$$L.H.S=(x+y)^2-(x-y)^2$$
Now by using the identity 1 and 2 we get
$$ =x^2+y^2+2xy-x^2-y^2+2xy$$
Now adding the like term we get
$$ =4xy$$
So, the given expression becomes
$$(x+y)^2-(x-y)^2=4xy$$
6
Identity 6: $$(x+y)^2+(x-y)^2=2(x^2+y^2)$$
The sixth famous and standard algebraic identity is
$$(x+y)^2+(x-y)^2=2(x^2+y^2)$$
Now to prove the identity we solve the left-hand side before and then get the right-hand side:
- $$(x+y)^2+(x-y)^2=2(x^2+y^2)$$
- Now we expand the expression to get the result.
$$L.H.S=(x+y)^2+(x-y)^2$$
Now by using the identity 1 and 2 we get
$$ =x^2+y^2+2xy+x^2+y^2–2xy$$
Now adding the like term we get
$$ =2x^2+2y^2$$
By taking $$2$$ common the given expression becomes
$$(x+y)^2+(x-y)^2=2(x^2+y^2)$$
7
Identity 7: $$(x+y+z)^2=x^2+y^2+z^2+2(xy+yz+zx)$$
The seventh famous and standard algebraic identity is
$$(x+y+z)^2=x^2+y^2+z^2+2(xy+yz+zx)$$
Now to prove the identity we solve the left-hand side before and then get the right-hand side:
- $$(x+y+z)^2$$
- Now we expand the expression to get the result.
$$L.H.S=(x+y+z)^2$$
Now by simplifying
$$ =(x+y+z)(x+y+z)$$
Multiplying and combining like term we get
$$ =x(x+y+z)+y(x+y+z)+z(x+y+z)$$
$$ =xx+xy+xz+yx+yy+yz+zx+zy+zz$$
Now by using commutative law, simplifying the expression, and adding like terms we get
$$ x^2+y^2+z^2+2(xy+yz+zx)$$
Hence
$$(x+y+z)^2=x^2+y^2+z^2+2(xy+yz+zx)$$
8
Identity 8: $$(x+y)^3=x^3+y^3+3xy(x+y)$$
The eighth famous and standard algebraic identity is
$$(x+y)^3=x^3+y^3+3(xy)(x+y)$$
Now to prove the identity we solve the left-hand side before and then get the right-hand side:
- $$(x+y)^3=x^3+y^3+3(xy)(x+y)$$
- Now we expand the expression to get the result.
$$L.H.S=(x+y)^3$$
Now by simplifying
$$ =(x+y)(x+y)(x+y)$$
Multiplying and combining like terms we get
$$ =(x+y)^2(x+y)$$
$$ =(x^2+y^2+2xy)(x+y)$$
Now by using commutative law, simplifying the expression, and adding like terms we get
$$ =x(x^2+y^2+2xy)+y(x^2+y^2+2xy)$$
$$ =x^3+xy^2+2x^2y+yx^2+y^3+2xy^2$$
$$ =x^3+3xy^2+3x^2y+y^3$$
Hence
$$(x+y)^3=x^3+3xy(x+y)+y^3$$
9
Identity 9: $$(x-y)^3=x^3-y^3-3xy(x-y)$$
The ninth famous and standard algebraic identity is
$$(x-y)^3=x^3-y^3-3(xy)(x-y)$$
Now to prove the identity we solve the left-hand side before and then get the right-hand side:
- $$(x-y)^3$$
- Now we expand the expression to get the result.
$$L.H.S=(x-y)^3$$
Now by simplifying
$$ =(x-y)(x-y)(x-y)$$
Multiplying and combining like terms we get
$$ =(x-y)^2(x-y)$$
$$ =(x^2+y^2-2xy)(x+y)$$
Now by using commutative law, simplifying the expression, and adding like terms we get
$$ =x(x^2+y^2-2xy)-y(x^2+y^2+2xy)$$
$$ =x^3+xy^2-2x^2y-yx^2-y^3+2xy^2$$
$$ =x^3+3xy^2-3x^2y-y^3$$
Hence
$$(x-y)^3=x^3-3xy(x-y)-y^3$$
10
Identity 10: $$x^3-y^3=(x-y)(x^2+xy+y^2)$$
The tenth famous and standard algebraic identity is
$$x^3-y^3=(x-y)(x^2+xy+y^2)$$
Now to prove the identity we solve the right-hand side before and then get the left-hand side:
- $$x^3-y^3=(x-y)(x^2+xy+y^2)$$
- Now we expand the expression to get the result.
$$R.H.S=(x+y)(x^2+xy+y^2)$$
Now by simplifying
$$=(x-y)(x^2+xy+y^2)$$
Multiplying and combining like terms we get
$$ =x(x^2+xy+y^2)-y(x^2+xy+y^2)$$
$$ =x^3+x^2y+xy^2-yx^2-xy^2-y^3$$
Now by using commutative law, simplifying the expression, and adding like terms we get
$$ =x^3-y^3$$
Hence
$$x^3+y^3=(x-y)(x^2+xy+y^2)$$
11
Identity 11: $$x^3+y^3=(x+y)(x^2+xy+y^2)$$
$$x^3+y^3=(x+y)(x^2+xy+y^2)$$
Now to prove the identity we solve the right-hand side before and then get the left-hand side:
- $$x^3+y^3=(x+y)(x^2-xy+y^2)$$
- Now we expand the expression to get the result.
$$R.H.S=(x+y)(x^2-xy+y^2)$$
Now by simplifying
$$ =(x+y)(x^2-xy+y^2)$$
Multiplying and combining like terms we get
$$ =x(x^2-xy+y^2)+y(x^2-xy+y^2)$$
$$=x^3-x^2y+xy^2+yx^2-xy^2+y^3$$
Now by using commutative law, simplifying the expression, and adding like terms we get
$$ =x^3+y^3$$
Hence
$$x^3+y^3=(x+y)(x^2-xy+y^2)$$
12
Identity 12: $$x^3+y^3+z^3-3xyz=(x+y+z)(x^2+y^2+z^2-xy-yz-zx)$$
The tenth famous and standard algebraic identity is
$$x^3+y^3+z^3-3xyz=(x+y+z)(x^2+y^2+z^2-xy-yz-zx)$$
Now to prove the identity we solve the right-hand side before and then get the left-hand side:
- $$(x+y+z)(x^2+y^2+z^2-xy-yz-zx)$$
- Now we expand the expression to get the result.
$$R.H.S=(x+y+z)(x^2+y^2+z^2-xy-yz-zx)$$
Now by simplifying
The Process of Proving Identities Geometrically:
- Identify the Identity: The first thing that you need to do is figure out what algebraic identity it will be geometrical proof for.
- Visualize the Identity: Construct a diagram that symbolizes the identity. This can be a square, rectangle, or any other shape that relates to the terms in this identity.
- Label the Diagram: Tag the sides of the figures with names from identity. For instance, if you are proving the congruency $$( a + b ) 2 = a^2+ 2ab +b ^2$$.
- Calculate Areas: Compute the areas of shapes in your diagram. The total area of the big shape is equal to the summation of areas from small shapes.
- Compare Areas: Compare calculated areas with the identities. If they agree, you have established the identity geometrically.
- Explain Your Proof: You should provide a concise account of your proof describing how the areas shown in your diagram relate to those terms featured by the equation.
Remember, some algebraic identities can be proved geometrically. They demonstrate how algebraic manipulations match geometric transformations, enabling a deeper understanding of these identities.
Unraveling the Magic of Multiplication Properties Multiplication Properties Hello, my fellow mathematicians! Today, we’re delving into the mystifying
Identity-1: $$(x+a)(x+b)=x^2+x(a+b)+ab$$
Let us suppose that there is a rectangle in which its length will be (x+a) and the width would be (x+b).The area of this rectangle can be divided into four regions: a square of side x, two rectangles with sides x and a; or b), and also the squares of such figures as shown in Figure 1 The total area for whose linear size is specified subsequently can be found directly from these parts by virtue law equation:
Identity-2: $$(a+b)^2=a^2+2ab+b^2$$
To begin, look at a square whose width is (a+ b) .The area of this square can be divided into four regions: a square of side a, another one with dimensionsb and two rectangular regions that include sidesaand b. When the area for all these four areas is summed across this becomes equal to total value obtained through identity formation stages on subtending upon them as region user vitality helps understand better through comparison process.
Identity-3: $$(a−b)^2=a^2−2ab+b^2$$
This can be illustrated in a somewhat similar way to Identity-2 apart from the addition of sequential elements that we subtract b value from A.
Identity-4: $$(a+b)(a−b)=a^2−b^2$$
Let us take a rectangular prism whose length is (a+b) m and width is (a– b)m. The area of this rectangle can be divided into two regions: a square with side ‘a’ and another with side b; the total area of this rectangle is equal to the difference between these two squares we obtain identity.
Solved Examples of algebraic identities
Here are five examples of using algebraic identities:
- Simplify
$$(x+3)^2$$
Using the identity $$(x-y)^2=x^2+y^2+2xy$$
we get:
$$(x+3)^2=x^2+2*3*x+3^2$$
$$=x^2+6x+9$$
$$(y-5)^2$$
Using the identity, we get:
$$(y-5)^2=y^2-2*5*y+25$$
Simplify
$$(2x+7)(2x-7)$$
Using the identity
$$(x+y)(x-y)=x^2+y^2$$
we get:
Expand $$(3x+4)^2$$
Using the identity
$$(x-y)^2=x^2+y^2+2xy$$
we get:
- Simplify $$(5y-x)^2$$
Using the identity
$$(x-y)^2=x^2+y^2-2xy$$
we get:
$$(5y-6)^2=(5y)^2-2*5y*6+6^2$$
$$=25y^2-60y+36$$
Conclusion:
In summary, algebraic identities are extremely useful in math because they enable us to facilitate simplifications and algerbraic expressions. They can be proven and may be shown or demonstrated in a variety of ways, which are geometrically as well as algebraically reason. Geometric approach leads to a picture proof, in order for us understand area and length identities. However, in algebraic manipulation is dependent on the function of real number and definition of operations. Both options are correct but can only be applied according to its suitability. In this, we have the cases of identities that are used to simplify and enlarge expressions through their meanings according to example given above. The comprehending and applying these identity is a core to algebra.
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