A viral post is being shared on social media which claims that police have started a 'free travel scheme' for women between 10 pm and 6 am. It stated two numbers - 1091 and 7837018555 as the helpline. An addition of two numbers is the same to the multiplication of those two numbers only when a,b = 0 or a,b = 2. (0+0=0*0 and 2+2=2*2) But since the first term is squared, (2+2)² - 2*2 = 16 - 4 â 0. This leaves a,b = 0 as the only option to get a zero. A second try was. a² + ab + b² = 0. The discriminant of a2 + a + 1 is negative and so you know the solutions will be complex non-real. As you need complex solutions for a, you should thus write it as a complex number: a = r[cos(θ) + isin(θ)]. Insert this in the equation a3 â 1 = 0 and find the solutions for r and θ. This yields: a3 â 1 = 0 â r3[cos(θ) + isin(θ)]3 â 1 If A³ is understood as the 3-fold cartesian product of A (and its members are 3-uples) and not the concatenation of three letters of alphabet A, then it's not exactly the Kleene star. But There seems to be no distinction made. The Binomial Series. This section looks at Binomial Theorem and Pascals Triangle. Pascal's Triangle. You should know that (a + b)² = a² + 2ab + b² and you should be able to work out that (a + b)³ = a³ + 3a²b + 3b²a + b³ . It should also be obvious to you that (a + b)¹ = a + b . This sequence is known as Pascal's triangle. Consider for example the n x n matrix A which is all zeros except for the top-right corner where it's 1, then A² = 0 and you don't get an n-dimensional space; in fact you get a 2 dimensional one. Such matrices are called nilpotent (of degree 2) and Similar stuff holds about nilpotent matrices of higher degrees (and those always exist: just a³ + b³ = (a + b)(a² â ab + b²) Sum of Two Cubes - Subtraction. a³ â b³ = (a â b)(a² + ab + b²) Factor by Grouping. ra + rb + sa + sb = r(a + b) + s(a + b) = (r +

s)(a + b) Divide in half, then factor what you can, take the common multiple and take the factors and put them into parentheses multiplied by each other. Factorise : a³ - 8b³ - 64c³ - 24abc. Summary: On factorising a³ - 8b³ - 64c³ - 24abc we get the factors as (a - 2b - 4c) (a² + 4b² + 16c² + 2ab - 8bc + 4ac. â Related Questions: Factorise : 2â2a³ + 8b³ - 27c³ + 18â2abc. Without actually calculating the cubes, find the value of : (1/2)³ + (1/3)³ - (5/6)³. Without actually Solution: Factorization of an algebraic expression refers to finding out the factors of the given algebraic expression. Given a³ + a² + a + 1. This expression does not have a common factor in all the terms. Rearranging the terms, we can write it as, a³ + a² + a + 1 = a² (a + 1) + a + 1. = a² (a + 1) + 1 (a + 1) NCERT Exemplar Class 9 Maths Exercise 2.3 Problem 34(ii) Factorise : a³ - 2â2b³. Summary: On factorising a³ - 2â2b³ using the algebraic identity we get (a - â2b)(a² + 2b² + â2ab) a² à a³ = Add the exponents together. Quotient Rule for Exponents a² ÷ a³ = a² over a³ = (top exponent) minus (bottom exponent) Rule for Negative Exponents aâ»â¿ = Reciprocal of the base. If the first term is of the form x² and the last term is of the form a², then the middle term must be of the form 2ax or â2ax. Terms in this set (14) Start studying algebraicke vzorce. Learn vocabulary, terms, and more with flashcards, games, and other study tools. If that's the case I set sqrt (3 2 + 1 2 + 0 2) = sqrt (0 2 + α 2 - 1 2). From that I get 10 = α 2 - 1 which leaves me with α = +/- sqrt (11). My textbook says the answer is +/- 3. Where has my understanding gone wrong? Thank you! Thanks! I hate when I get stuck over a careless mistake. If two vectors have the same length, then they're Find the lowest common denominator for the fractions shown.

3a^2/a^2+a+1, 2a/a-1, a^3/a^3-1. Log in Sign up. Find A Tutor . Search For Tutors. Request A Tutor. Online Tutoring. How It Works . For Students. FAQ. What Customers Say. Resources . Ask An Expert. Search Questions. Ask a Question. Lessons. Wyzant Blog. Start Tutoring . To simplify the expression (2a³ + 3a² + 7a) + (a³ + a² - 2a), you combine like terms to get the final result of 3a³ + 4a² + 5a. Explanation: To combine like terms in the algebraic expression (2a³ + 3a² + 7a) + (a³ + a² - 2a), we add the coefficients of the corresponding powers of a. For the cube terms, we have 2a³ and a³, which sum When as students we started learning mathematics, it was all about natural numbers, whole numbers, integrals. Then we started learning about mathematical functions like addition, subtraction, BODMAS and so on. Suddenly from class 8 onwards mathematics had alphabets and letters! Today, we will focus on algebra formula. (a) 3a (a² + 2) (b) 3 (a³ + 2) (c) a (3a² + 6) (d) 3 à a à a à a + 2 à 3 à a. Solution: The correct answer is (a), because. From the given expression 3a³ + 6a taking '3a' as a common factor we get 3a (a² + 2). Further we may notice that the term (a² + 2) is not having any term common and not factorable further. Transcribed Image Text: 7:58 个 Step3 c) possible minimal polynomial λ, A², A³ -3 1 1' A-6 2 2 3 1 1 A #0 A is not minimal -3 1 1 A² = -6 22 -3 1 1 = polynomial ÐÐ Ð Ð 000 000 A² is minimal polynomial -3 1 17 -6 22 -3 1 1 No Was this solution helpful? Yes âx DO. Expert Solution. Summary: By using the algebraic identity, it is proven that (a + b + c)³ - a³ - b³ - c³ = 3 (a + b) (b + c) (a + c) â Related Questions: Multiply x² + 4y² + z² + 2xy + xz - 2yz by (- z + x - 2y) If a, b, c are all non-zero and a + b + c = 0, prove that a²/bc + b²/ca + c²/ab = 3. If a + b + c = 5 and ab + bc + ca = 10, then prove A

expressão é equivalente a 4ab. Produtos notáveis. A expressão é: (a + b)² - (a - b)² - (a + b). (a - b) + a² - b² . Vamos separar os produtos notáveis: (a + b)²: Essa é uma fórmula de quadrado perfeito, que pode ser simplificada para a² + 2ab + b². (a - b)²: Essa também é uma fórmula de quadrado perfeito, que pode ser simplificada para a² - 2ab + b². NCERT Exemplar Class 8 Maths Chapter 7 Problem 113(ii) Verify: (a + b + c) (a² + b² + c² - ab - bc - ca) = a³ + b³+ c³ - 3abc. Summary: (a + b + c) (a² + b² + c² - ab - bc - ca) = a³ + b³+ c³ - 3abc is true