Weba3 − b3 a 3 - b 3 Since both terms are perfect cubes, factor using the difference of cubes formula, a3 −b3 = (a−b)(a2 + ab+b2) a 3 - b 3 = ( a - b) ( a 2 + a b + b 2) where a = a a = … WebApr 1, 2024 · 3(b −c)(c −a)(a − b). Explanation: If we use the following Result, we can immediately factorise the given Exp. = 3(b − c)(c − a)(a −b). Result : x + y + z = 0 ⇒ x3 …
Factorise: {a-b]3 + {b-c}3 + {c-a}3 Polynomials-Maths-Class-9
WebJul 30, 2024 · Factorisation Factorise (a+b)^3-8(a-b)^3. Factorisation class 9. #factorisation #factorisationavijainclasses #avijainclasses #polynomials #polynomialsclass... WebJun 4, 2014 · Factorise : a 3 (b - c) 3 + b 3 (c - a) 3 + c 3 (a - b) 3. Asked by Topperlearning User 04 Jun, 2014, 01:23: PM ... factorise the following a^3+b^3+3a-3b. Asked by s1044kasturipriyambada026582 27 May, 2024, 12:02: PM. ANSWERED BY EXPERT. CBSE 9 - Maths. I want to solve this question. rocker snowboard burton blender
Factoring by grouping (article) Khan Academy
WebStep 1: Factorize the given expression a 3 + b 3 + c 3 - 3 a b c use the identity ( x + y) 3 = x 3 + y 3 + 3 x y ( x + y) ⇒ ( x + y) 3 - 3 x y ( x + y) = x 3 + y 3 put x = b, y = c ∴ a 3 + b 3 + … WebAnswer: ab^3 - ac^3 + bc^3 - a^3b + a^3c - b^3c => rewrite as: = ab^3 - ac^3 - a^3b + a^3c - b^3c + bc^3 => factoring by grouping: = a(b^3 - c^3) - a^3(b - c) - bc(b^2 - c^2) => factor 1st and 3rd terms: = a(b - c)(b^2 + bc + c^2) - a^3(b - c) - bc(b - c)(b + c) => common factor b - c: = (b -... otc 61217