We know that a^3+b^3+c^3−3abc=(a+b+c)(a^2+b^2+c^2−ab−bc−ca) i.e. , =>(a+b)^3+(b+c)^3+(c+a)^3−3(a+b)(b+c)(c+a)=((a+b)+(b+c)+(c+a))[(a+b)^2+(b+c)^2+(c+a)^2−(a+b)(b+c)−(b+c)(c+a)−(c+a)(a+b)] =>(a...
Factorise completely : a^(2) + 2ab + b^(2) -c^(2) 00:47 Factorise completely : x^(2) + 6xy + 9y^(2) + x + 3y 02:00 Factorise completely : 4a^(2) - 12 ab + 9b^(2) + 4a - 6b 02:16 Factorise completely : 2a^(2) b^(2) - 98 b^(4) 01:20 Factorise completely...
百度试题 结果1 题目Fully factorise (vmatrix) a&1&a^3 b&1&b^3 c&1&c^3(vmatrix) 相关知识点: 试题来源: 解析 -(a-b)(b-c)(c-a)(a+b+c) 反馈 收藏
5 (a) Factorise 5a-3a2a(5-3a)2)(b)Expand(i)2(4-3w)8-6w(ii) 1^2(y+10)+10g(3)(e) W=(5.6
II.Fill in the blanks3Factorise:ab+62-ac-h =()-(ac +c)=()().4Factorise:ax2+ax-b-bx =(
Factorise the following by using suitable identities (a b) 2 (b c) 2 - Given:$(a-b)^2- (b-c)^2$To do:We have to factorise the given expression using suitable identities.Solution:We know that,$x^2-y^2=(x-y)(x+y)$Therefore,$(a-b)^2- (b-c)^2=[(a-b)-(b-c)...
Factorise: (9a+b)^2-9 01:16 Factorise: x-64x^3. 00:58 Factorise: x^2-32xy-105y^2 03:45 Factorise: x^2 +21x+108 04:28 Factorise: 6x^2+7x-3 03:41 Factorise: 42 x^2+10x-12 02:59 Factorise: 12y^2y-6 00:32 Factorise: x^2-10x+21. 02:30 Carry out the following divis...
To factorize the expression (a+b)3+(b+c)3+(c+a)3−3(a+b)(b+c)(c+a), we can use a known algebraic identity. 1. Recognize the Identity: The expression can be simplified using the identity: A3+B3+C3−3ABC=(A+B+C)(A2+B2+C2−AB−AC−BC) where A=a+b, B=b+...
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Step by step video, text & image solution for .Factorise: (2a-b-c)^(3)+(2b-c-a)^(3)+(2c-a-b)^(3) by Maths experts to help you in doubts & scoring excellent marks in Class 9 exams. Updated on:21/07/2023 Class 9MATHSPOLYNOMIALS ...