Algebraic Identities: Proofs, Properties and Examples

Algebra unifies all of mathematics. From solving a first-order polynomial equation to deriving equations of field theory, it is a handy tool. So, it becomes essential to have a correct grasp of the elementary algebraic concepts and techniques. The elementary algebra in its true sense translates an arithmetic equation from a set of unknowns. Though algebra is in itself a broad area of mathematics, which you must have learned in algebra’s introductory article, here the emphasis has been given to elementary algebra and its identities. Moreover, this article covers proofs of those identities and their applications to solve problems.

Key Takeaways: Algebra, identity, mathematics, variables, algebraic identities, cube, square, sum, difference.

Read More: Pair of Linear Equations in Two Variables Formula

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Algebraic Identities 

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Algebraic identities are a set of algebraic equations that hold true for all values of the variables. Some identities are trivial equations and using them provides a faster approach in solving problems and finding factors of a polynomial equation. Basic identities include numbers, unknowns or variables, and mathematical operators( multiplication, addition, division, and subtraction). 

Although algebraic identities are algebraic equations, all algebraic equations are not identities. For example, x - 5 = 10, or x = 15 is an algebraic equation, because the equation is true for only a certain value. Whereas, 5x+x=6x is an identity as the equation is true for all values of x. Now we proceed to mention the basic algebraic identities.

Read More: Coordinate Geometry

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Standard Algebraic identities

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Here are some standard algebraic identities and their proofs. Some of these were present in your previous class.

Identity I: Square of the sum of two terms

( a + b )² = a² + 2ab + b²

Proof: 

Method I

Consider a square with side ( a + b ) units

We divide the square into four quadrilaterals – two rectangles with sides ( a and b ) and two squares with sides a and b.

Therefore using area of square formula (side square) and area of rectangle (length* breadth), we get,

The area of the square with side (a + b) units is a² + ab + ab + b²

Therefore, (a + b)²= a² + 2ab + b²

Method II

Or, LHS = (a + b) (a + b)= a(a + b) + b(a + b) and RHS = a²+ 2ab + b²

Or, LHS = aa + ab + ba + bb = a² + 2ab + b²

Therefore LHS = RHS

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Identity II: Square of subtraction of two terms

( a - b )² = a² - 2ab + b²

Proof: 

Method I 

Consider a square with side a = ( a - b ) + b units

We divide the square into three quadrilaterals – two rectangles with sides and a square with sides ( a - b ).

Area of square with side a is a² = ( a-b )² + b ( a - b) + ab

Or, a² = ( a-b )² + b a - bb + ab

Or, a² = ( a-b )² + 2ab - b²

Or, ( a-b )² = a² - 2ab+ b²

Method II

Or, LHS = (a - b) (a - b)= a (a - b) - b (a - b) and RHS = a² - 2ab + b²

Or, LHS = aa - ab- ba + bb = a² - 2ab + b²

Therefore LHS = RHS

Read More: Elimination Method of Solving a Pair of Linear Equations 

Identity III: Difference of square of two terms

( a + b ) ( a - b )= a²- b²

Proof: 

Method I

Consider a square with side a = ( a - b ) + b units

We divide the square into three quadrilaterals – two rectangles with sides and a square with sides b .

From the figure, the total area of the square with side a can be given by, 

Area of square with side a is a² = a ( a-b ) + b ( a - b) + b²

Taking ( a-b ) common in the equation we get,

Or, a²  = ( a + b ) ( a - b) + b²

Or, a²- b²= ( a + b ) ( a - b)

Method II

Or, RHS = ( a + b ) ( a - b ) and LHS = a²- b²

Or, RHS = a ( a-b ) + b ( a - b )

Or, RHS = a a-a b + b a - b b

Or, RHS = a²- b²

Therefore LHS = RHS

Read More: Inverse Matrix Formula

Identity IV: Multiplication of two algebraic sums

( x + a ) ( x + b ) = x² + ( a + b ) x + ab

Proof: 

Method I

Consider a rectangle with length ( x + b ) units and breadth ( x + a ) units.

We divide the rectangle into four quadrilaterals – three rectangles with sides ( x, a ) , ( a, b ), and ( x, b ) . and a square of side x

Or, (x + a ) (x + b) = x² + xa + xb +ab

Or, (x + a ) (x + b) = x² + x (a + b ) +ab

Method II

LHS = (x + a ) (x + b) and RHS= x² + 2 (a+b) x + ab

Or, LHS = x (x + b) + a (x + b)

Or, LHS = xx +xb + a x +a b

Or, LHS= x² + (a+b) x + ab

Therefore LHS = RHS

Read More:

Identity V: Square of sum of three terms

a + b + c ² = a²+ b²+ c² +2ab + 2bc + 2ac

Proof: This identity can solved using ( b + c ) = B

Then the LHS can be written as a + b + c = ( a + B )

But we know (a + b)²=a² + 2ab + b² [ Identity I ] , we get ….(i)

Or, ( a + B )² =a² + 2aB + B²

Again using the identity I we get,

Or, ( a + B )² =a² + 2aB + B² = a² + 2a ( b + c ) + (b + c)²

Or, a + b + c² = a² + 2a ( b + c ) + ( b² + 2bc + c² )

Or, a + b + c²= a²+ b²+ c² +2ab + 2bc + 2ac = RHS

Identity VI: Cube of sum of two terms 

a + b ³ = a³ + b³+ 3ab ( a + b )

Proof: a + b³ = [ (a + b)² ] (a + b)

Or, a + b³ = [ a² + 2ab + b² ] (a + b) [ Using Identity I ]

Or, a + b³ = ( a² + 2ab + b² ) a + ( a² + 2ab + b² ) b

Or, a + b³ =a³ + b³+ 3ab (a+b)= RHS

Read More: Rolle’s Theorem

Identity VII: Cube of difference of two terms 

a - b ³= a³- b³- 3ab ( a - b )

Proof: a - b³ = [ (a - b)² ] (a - b)

Or, a - b³ = [ a² - 2ab + b² ] (a - b) [ Using Identity II ]

Or, a - b³ = ( a² - 2ab + b² ) a - ( a² - 2ab + b² ) b

Or, a - b³ =a³ - b³ - 3ab (a - b)= RHS

Identity VIII: Sum of Cubes

a³+b³ = ( a + b ) ( a²- ab + b² )

Proof: Using Identity VI, we get –

Or, a + b ³ = a³ + b³+ 3ab (a+b)

Or, a³ + b³+ 3ab ( a + b ) = a + b³

Or, a³ + b³ = a + b 3 - 3ab ( a + b )

Or, a³ + b³ = a + b [ ( a + b )²  - 3ab ]

Or, a³ + b³ = a + b ( a² + 2ab + b²- 3ab )

Or, a³ + b³ = ( a + b ) ( a²- ab + b²)

Read More: Complex Numbers and Quadratic Equations

Identity IX: Difference of Cubes

a³-b³ = ( a - b ) ( a² + ab + b²)

Proof: Using Identity VII, we get –

Or, a - b³ =a³ - b³- 3ab ( a - b )

Or, a³ - b³- 3ab ( a - b ) = a - b ³

Or, a³ - b³ = a - b³ + 3ab ( a - b )

Or, a³ - b³ = a - b[( a - b )²  + 3ab ]

Or, a³ - b³= a - b (a² - 2ab + b²+ 3ab )

Or, a³ - b³ = (a - b) (a² + ab + b²)

Read More: Maxima Minima

Identity X: Sum of cubes of three terms and their product

a³ + b³ + c³– 3abc = ( a + b + c ) ( a² + b² + c² – ab – bc – ca )

Proof: 

Consider the RHS ( a + b + c ) ( a² + b² + c2 – ab – bc – ca )

Or, RHS = a ( a² + b² + c² – ab – bc – ca ) + b ( a² + b² + c² – ab – bc – ca ) + c ( a² + b² + c² – ab – bc – ca )

Or, RHS = = ( a³ + a b² + a c² - ba²- abc - ca² ) + ( ba²+ b³+ b c² - ab²- abc - cb² ) + ( ca²+ c b²+ c³ - bc²- abc - ac² )

Combining like terms, we get

RHS = a³ + b³ + c³ – 3abc

Therefore, we get the identity

a³ + b³ + c³ – 3abc = ( a + b + c ) ( a² + b² + c² – ab – bc – ca )

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Things to Remember

• An identity is an equality that is true for all values. However, algebraic identities are algebraic equations but all algebraic equations are not identities. The value of an algebraic expression will change if the values of variables are changed.
• Some useful identities are –
  1. ( a + b )²=a² + 2ab + b²
  2. ( a - b )² = a² - 2ab + b²
  3. (a+b) (a-b)= a²- b²
  4. ( x + a ) ( x + b ) = x² + ( a + b ) x + ab
  5. a + b + c ² = a²+ b²+ c² +2ab + 2bc + 2ac
  6. a + b ³ = a³ + b³+ 3ab ( a + b )
  7. a - b ³= a³- b³- 3ab ( a - b )
  8. a³+b³ = ( a + b ) ( a²- ab + b² )
  9. a³-b³ = ( a - b ) ( a² + ab + b²)
  10. a³ + b³ + c³ – 3abc = ( a + b + c ) ( a² + b² + c² – ab – bc – ca )
• These identities can be verified either using a substitution method ( substitute the values for the variables and perform the arithmetic operations ) or using simple geometrical models.