Math, asked by rajagoud22, 11 months ago

write all the identities of aldebra​

Answers

Answered by BrainlyRaaz
10

Algebra

(i) Expansions :

(a ± b)²

(a ± b)³

(x ± a) (x ± b)

(ii) Factorisation

a² - b²

a³ ± b³

ax² + bx + c, by splitting the middle term.

(iii) Changing the subject of a formula.

  • Concept that each formula is a perfect equation with variables.

  • Concept of expressing one variable in terms of another, various operators on terms - transposing the terms squaring or taking square root etc.

(iv) Linear Equations and Simultaneous (Linear) Equations

(a). Solving algebraically (by elimination as well as substitution) and graphically.

(b) Solving simple problems based on these by framing appropriate formulae.

(v) Indices / Exponents

Handling positive, fractional, negative and "zero" indices.

Simplification of expressions involving various exponents

a^m x a^n = a^(m + n) , a^m ÷ a^n = a^(m-n), (a^m)^n = a^(mn) etc, use of laws of exponents.

(vi) Logarithms

(a) Logarithmic form vis - a - vis exponential form: interchanging.

(b) Laws of Logarithms and its use

Expansion of expression with the help of lacws of logarithm

e. g. y = a⁴ × b²/c³

log y = 4 log a + 2 log b - 3 log c etc.

Some Standard Algebraic Identities list are given below:

Identity I: (a + b)^2 = a^2 + 2ab + b^2

Identity II: (a – b)^2 = a^2 – 2ab + b^2

Identity III: a^2 – b^2= (a + b)(a – b)

Identity IV: (x + a)(x + b) = x^2 + (a + b) x + ab

Identity V: (a + b + c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ca

Identity VI: (a + b)^3 = a^3 + b^3 + 3ab (a + b)

Identity VII: (a – b)^3 = a^3 –b^3–3ab (a – b)

Identity VIII: a^3 + b^3 + c^3 – 3abc = (a + b + c)(a^2 + b^2 + c^2 – ab – bc – ca)

Answered by Anonymous
11

Answer:

Algebraic identities:

★ (a + b)² = a² + b² + 2ab

★ (a - b)² = a² + b² - 2ab

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

★ (a² - b²) = (a + b)(a - b)

★ (a + b)³ = a³ + b³ + 3ab(a + b)

★ (a - b)³ = a³ - b³ - 3ab(a - b)

★ a³ + b³ = (a + b)(a² + b² - ab)

★ a³ - b³ = (a - b)(a² + b² + ab)

★ a³ + b³ + c³ - 3abc = (a + b + c)(a² + b² + c² - ab - bc - ac)

★ if (a + b + c) = 0 then, a³ + b³ + c³ = 3abc

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