**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.**

Definition: An identity is an equality, which is true for all values of the variables in the equality.

Note: An equation is true for only certain values of the variable in it. It is not true for all values of the variable.

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**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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