Factorise: (a – b)3 + (b – c)3 + (c – a)3

Asked by Last Modified  

12 Answers

Learn Mathematics +1

Follow 15
Answer

Please enter your answer

Krishna Yeswanth

3(a-b)(b-c)(c-a)
1 Comments
Dislike Bookmark

3
Comments
Dislike Bookmark

Math Tutor

3(a-b+b-c+c-a)= 3*(0)=0
Comments
Dislike Bookmark

Tutor

First factorize (a-b)^3 , (b-c)^3 and (c-a)^3 separately and then add them together to get the final result. (a-b)^3 = a^3 - b^3 - 3a^2b + 3ab^2 ................(1) (b-c)^3 = b^3 - c^3 - 3b^2c + 3bc^2 .................(2) (c-a)^3 = c^3 - a^3 - 3ac^2 + 3 a^2c ...
read more
First factorize (a-b)^3 , (b-c)^3 and (c-a)^3 separately and then add them together to get the final result. (a-b)^3 = a^3 - b^3 - 3a^2b + 3ab^2 ................(1) (b-c)^3 = b^3 - c^3 - 3b^2c + 3bc^2 .................(2) (c-a)^3 = c^3 - a^3 - 3ac^2 + 3 a^2c ..................(3) Now add (1), (2) and (3). This gives, a^3 - b^3 - 3a^2b + 3ab^2 + b^3 - c^3 - 3b^2c + 3bc^2 + c^3 - a^3 - 3ac^2 + 3 a^2c = -3a^2b + 3ab^2 - 3b^2c + 3bc^2 - 3ac^2 + 3 a^2c read less
Comments
Dislike Bookmark

Tutor

a3-3a2b+3ab2-b3 + b3-3b2c+3bc2-c3 + c3-3c2a+3ca2-a3 = -3a2b+3ab2 -3b2c+3bc2 -3c2a+3ca2 = 3a2 ( c-b) + 3b2 ( a-c ) + 3c2 (b-a)
Comments
Dislike Bookmark

Math Magician

(a-b)(b-c)(c-a)
Comments
Dislike Bookmark

Maths Trainer with 23 years of experience

we know, x3+y3=(x+y)(x2−xy+y2)x3+y3=(x+y)(x2−xy+y2) so, (a−b)3+(b−c)3=(a−c)((a−b)2−(a−b)(b−c)+(b−c)2) (a−b)3+(b−c)3=(a−c)((a−b)2−(a−b)(b−c)+(b−c)2) now, (a−b)3+(b−c)3+(c−a)3=(a−c)((a−b)2−(a−b)(b−c)+(b−c)2)+(c−a)3=(c−a)(−(a−b)2+(a−b)(b−c)−(b−c)2+(c−a)2) (a−b)3+(b−c)3+(c−a)3=(a−c)((a−b)2−(a−b)(b−c)+(b−c)2)+(c−a)3=(c−a)(−(a−b)2+(a−b)(b−c)−(b−c)2+(c−a)2) now,...
read more
we know, x3+y3=(x+y)(x2−xy+y2)x3+y3=(x+y)(x2−xy+y2) so, (a−b)3+(b−c)3=(a−c)((a−b)2−(a−b)(b−c)+(b−c)2) (a−b)3+(b−c)3=(a−c)((a−b)2−(a−b)(b−c)+(b−c)2) now, (a−b)3+(b−c)3+(c−a)3=(a−c)((a−b)2−(a−b)(b−c)+(b−c)2)+(c−a)3=(c−a)(−(a−b)2+(a−b)(b−c)−(b−c)2+(c−a)2) (a−b)3+(b−c)3+(c−a)3=(a−c)((a−b)2−(a−b)(b−c)+(b−c)2)+(c−a)3=(c−a)(−(a−b)2+(a−b)(b−c)−(b−c)2+(c−a)2) now, (c−a)2−(a−b)2=(c−a+a−b)(c−a−a+b)=(c−b)(c−2a+b)(c−a)2−(a−b)2=(c−a+a−b)(c−a−a+b)=(c−b)(c−2a+b) the expression becomes, (c−a)((c−b)(c−2a+b)+(b−c)(a−2b+c))=(c−a)(b−c)(−c+2a−b+a−2b+c) =3(c−a)(b−c)(a−b) read less
Comments
Dislike Bookmark

Teacher

ans is zero
Comments
Dislike Bookmark

Tutor

We know a^3 + b^3 +c^3= 3abc if a+b+c=0...so answer is 3(a-b)(b-c)(c-a)
Comments
Dislike Bookmark

3(a-b)(b-c)(c-a)
Comments
Dislike Bookmark

View 10 more Answers

Related Questions

What is the value of Sin360+Cos360?
Sin(360)= sin(0)=0 Cos(360)=cos(0)=1 Formula: Sin((360*x)+y)=sin(y) Cos((360*x)+y)=cos(y) Where x is any integer and y is any real number. Sin(360) +cos (360) =1
Ramana
How much monthly fee for class 10 for 1 subject like Maths in Delhi or NCR?
The best way is to charge as per hour. 300/- per hour will be considered acceptable for parents and you.
Vishal Ranjan
What is a closed figure formed by three line segments called?
Triangle
Sense Softech
0 0
7

Hello everyone

Which is better? : Online tuitions by brands like: Meritnation, Myschool page or Catalyze or regular tuition classes?
I would suggest offline coaching because it binds the students and gives different levels of class surrounding, which is not possible in online classes. However, for extra self-studies, one can take the help of online material.
Swetha
A triangular park has sides 120 m, 80 m and 50 m. A gardener has to put a fence all around it and also plant grass inside. How much area does he need to plant? Find the cost of fencing it with barbed wire at the rate of Rs. 20 per metre, leaving a space of 3 m wide for a gate on one side.
using Henon's(s-a)(s-b)(s-c)=?125×5×45×75=(25×15)?15=375?15sq.m Cost of fence without gate=(250-3)×20=247×20=RS.4940
Princy
0 0
5

Now ask question in any of the 1000+ Categories, and get Answers from Tutors and Trainers on UrbanPro.com

Ask a Question

Related Lessons

Lokesh Joshi

0 0
0

Lokesh Joshi

0 0
0
Class-9-Maths L-11 Surface Area & Volume Ex 11.1 Q3

Ashish Bhatia

0 0
0
Class-9-Maths L-11 Surface Area & Volume Ex 11.4 Q4

Ashish Bhatia

0 0
0

Lokesh Joshi

0 0
0

Recommended Articles

Top Benefits of e-Learning

With the current trend of the world going digital, electronic renaissance is a new movement that is welcomed by the new generation as it helps makes the lives of millions of people easier and convenient. Along with this rapidly changing movement and gaining popularity of Internet, e-Learning is a new tool that emerging...

Read full article >

How To Choose The Right School For Children?

Quality education does not only help children to get a successful career and life, but it also hugely contributes to society. The formal education of every child starts from school. Although there are numerous schools, parents find it challenging to choose the right one that would fit their child. It is difficult for them...

Read full article >

The Rising Problems in Indian Government School

With the mushrooming of international and private schools, it may seem that the education system of India is healthy. In reality, only 29% of children are sent to the private schools, while the remaining head for government or state funded education. So, to check the reality of Indian education system it is better to look...

Read full article >

The Trend of E-learning and it’s Benefits

E-learning is not just about delivering lessons online. It has a much broader scope that goes beyond manual paper or PowerPoint Presentations. To understand the reach of E-learning and how the whole process works in developing the Educational system, we will discuss a few points here. Let us find out how this new learning...

Read full article >

Find Class 9 Tuition near you
Class 9 Tuition in Delhi
Class 9 Tuition in Bangalore
Class 9 Tuition in Hyderabad
Class 9 Tuition in Kolkata
Class 9 Tuition in Mumbai
Class 9 Tuition in Chennai
Class 9 Tuition in Pune
Class 9 Tuition in Noida
Class 9 Tuition in Gurgaon
Class 9 Tuition in Lucknow

Looking for Class 9 Tuition ?

Learn from the Best Tutors on UrbanPro

Are you a Tutor or Training Institute?

Join UrbanPro Today to find students near you