Algebraic Identities

Basic and Advanced Algebraic Identities

What are Algebraic identities?

Let us understand this by using one of the examples from Algebraic Identities Class 8 NCERT Book.

Consider this Example: (a + 1) (a +2) = a² + 3a + 2, and Find the value for a = 2 and a = 3.

For a = 2

LHS = (a+1)(a+2) = (2+1)(2+2) = 3×4 = 12

RHS = a² + 3a + 2 = 2² + 3×2 + 2 = 4 + 6 + 2 = 12

Therefore, for a = 2, LHS=RHS

Similarly, for a = 3

LHS = (a+1)(a+2) = (3+1)(3+2) = 4×5 = 20

RHS = a² + 3a + 2 = 3² + 3×3 + 2 = 9 + 9 + 2 = 20

For a = 3, LHS=RHS

From the above two values of ‘a’ we see that the LHS=RHS, and hence we can say if an equality is true for every value of the variable in it, is called an identity.

Algebraic Identities help us to solve the algebraic expressions or problems related to algebra easily and in less time. Algebraic Identities also help in factorization of an algebraic expression. 

Profit And Loss Important Formulas

Algebraic Identities Class 8

(a + b)² = a² + 2ab + b²

(a – b)² = a² – 2ab + b²

(a + b) (a – b) = a² – b²

These above three identities are called the Standard Identities.

(x + a) (x + b) = x² + (a + b) x + ab

Algebraic Identities Class 9

(x + y + z)² = x² + y² + z² + 2xy + 2yz + 2zx

(x – y – z)² = x² + y² + z² – 2(xy + yz – zx)

(x + y)³ = x³ + y³ + 3xy (x + y)

(x – y)³ = x³ + y³ – 3xy (x – y)

(x – y)³ = x³ + y³ – 3x²y + – 3xy²

x³ + y³ + z³ – 3xyz = (x + y + z) (x² + y² + z² – xy – yz – zx)

If x + y + z = 0, then x³ + y³ + z³ = 3xyz

Gravitation Class 9 Numerical with Solution

Algebraic Identities Class 10

x³ + y³ = (x + y)³ – 3xy (x + y)

x³ + y³ = (x + y) (x² -xy + y²)

x³ – y³ = (x – y)³ + 3xy (x – y)

x³ + y³ = (x – y) (x² + xy + y²)

(a + b + c)³ = a³ + b³ + c³ + 3(a + b)(b + c)(c + a)

Extra Formula of Algebraic Identities

These formulas will help you in solving Algebraic Identities questions with an extra edge or the advanced level questions that often come in the SSC CGL, SSC CHSL, Banking and other government Exams.

 a² – b² = (a + b) (a – b)

 a² + b² = (a + b)² -2ab

a+b=\sqrt{{ \left( a-b \right)}^{2}+4ab} a-b=\sqrt{{ \left( a+b \right)}^{2}-4ab}

If x+\frac { 1 }{ x } \quad =\quad k

then { x }^{ 2 }+\frac { 1 }{ { x }^{ 2 } } ={ k }^{ 2 }-2

If x-\frac { 1 }{ x } \quad =\quad k

then { x }^{ 2 }+\frac { 1 }{ { x }^{ 2 } } ={ k }^{ 2 }+2

If x+\frac { 1 }{ x } \quad =\quad k

then { x }^{ 3 }+\frac { 1 }{ { x }^{ 3 } } ={ k }^{ 3 }-3k

If x-\frac { 1 }{ x } \quad =\quad k

then { x }^{ 3 }-\frac { 1 }{ { x }^{ 3 } } ={ k }^{ 3 }+3k

If { x }^{ 2 }+\frac { 1 }{ { x }^{ 2 } } =k

then { x }+\frac { 1 }{ { x } } =\sqrt { k+2 }

If { x }^{ 2 }+\frac { 1 }{ { x }^{ 2 } } =k

then { x }-\frac { 1 }{ { x } } =\sqrt { k-2 }

Algebraic Identities Examples

Example 1 based on Algebraic Identities (a + b)² = a² + 2ab + b²

Question: Find the value of 103².

(103)² = (100 + 3)² = 100² + 2 × 100 × 3 + 3² = 10000 + 600 + 9 = 10609 [Answer]

Example 2 based on Algebraic Identities (a – b)² = a² – 2ab + b²

Question: Find the value of 4.9².

(4.9)² =(5.0 – 0.1)² = (5.0)² – 2 (5.0) (0.1) + (0.1)² = 25.00 – 1.00 + 0.01 = 24.01

Example 3 based on Algebraic Identities (a + b) (a – b) = a² – b²

Question: Find the value of 194 × 206

194 × 206 = (200 – 6) × (200 + 6) = 200² – 6² = 40000 – 36 = 39964

Example 4 based on Algebraic Identities (x + a) (x + b) = x² + (a + b) x + ab

Question: Find the value of 95 × 103

95 × 103 = (100 – 5) × (100 + 3) = 1002 + (–5 + 3) × 100 + (–5) × 3 = 10000 – 200 – 15 = 9785

Example 5 based on Algebraic Identities (x + y + z)² = x² + y² + z² + 2xy + 2yz + 2zx

Question: Write (3a + 4b + 5c)² in expanded form.

(3a + 4b + 5c)² = (3a)² + (4b)² + (5c)² + 2(3a)(4b) + 2(4b)(5c) + 2(5c)(3a)

= 9a² + 16b² + 25c² + 24ab + 40bc + 30ac

Example 6 based on Algebraic Identities (x + y)³ = x³ + y³ + 3xy (x + y)

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