  Question: Factorize : 3(a+b) + 3(b+c) + (c+a)3

Posted by: Kevin on 12.02.2021

K
Kameshwar R. from Visakhapatnam
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3(a+b)+3(b+c)+3(c+a) = 3a+3b+3b+3c+3c+3a =6a+6b+6c =6(a+b+c) Shubham B. from Kolkata
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3(a+b) + 3(b+c) + 3(c+a) = 3a+3b+3b+3c+3c+3a =6a+6b+6c =6(a+b+c). Answer Deepak K. from Bhopal
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6(a+b+c)
V
Vanitha N. from Bangalore
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3(a+b)+3(b+c)+(c+a)3 =3a+3b+3b+3c+3c+3a =6a+6b+6c =6(a+b+c) Ranjan S. from Delhi
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3(a+b)+3(b+c)+3(c+a) After open bracket 3a+3b+3b+3c+3c+3a Adding similar terms 6a+6b+6c = 6(a+b+c) answer Karamjeet S. from Delhi
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3(a+b)+3(b+c)+3(c+a)= 3a+3b+3b+3c+3c+3a= 6a+6b+6c= 6(a+b+C) Ans Rathindra N. from Kolkata
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3(a+b)+3(b+c)+(c+a)3 =3a+3b+3b+3c+3c+3a =6a+6b+6c 6(a+b+c) Swapnil C. from Pune
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= 3(a+b)+3(b+c)+3(c+a) By taking 3 common we get, = 3(a+b+b+c+c+a) = 3×(2a+2b+2c) By taking 2 common from bracket we get, = 3×2×(a+b+c). Which is required factorization.
N
Neeraj C. from Delhi
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3(a+b)+3(b+c)+(c+a)3

answer= take 3 common, we get

= 3(a+b+b+c+c+a)

add like pair in bracket we get

= 3(2a+2b+2c)

take 2 common

= 6(a+b+c)

S
Shilja from Bangalore
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3a+3a+3b+3b+3c+3c= 6a+6b+6c= 6(a+b+c) Rathindra N. from Kolkata
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6(a+b+c)
M
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3(a+a)+3(b+b)+3(c+c) 3(2a)+3(2b)+3(2c) 6a+6b+6c 6(a+b+c) Pankaj S. from Delhi
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3(a+b) +3(b+c) +3(c+a) 3a + 3b +3b + 3c + 3c +3a 6a + 6b + 6c 6(a+b+c)  Chhaya R. from Noida
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First of all, multiply 3 with each bracket then add the same values to each other, take out common value from them.  Manju W. from Bangalore
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Step1-First multiply 3 to each bracket 3a+3b+3b+3c+3c+3a Step2-collect the terms of same variable together 3a+3a+3b+3b+3c+3c Step3-add terms of same variable 6a+6b+6c Step4-take out common 6 6(a+b+c)..
H
Himanshi T. from Delhi
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=3a+3b+3b+3c+3c+3a =6a+6b+6c =6(a+b+c)
U
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3a+3b+3b+3c+3c+3a= 6a+6b+6c=6(a+b+c)
S
Suman A. from Kolkata
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6( a + b + c )

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