$72$
$88$
$96$
$80$

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my answer by simple rule of inspection a=-1 b=+4 c=-3 reversible for values of a, b , c so (a+b)(b+c)(c+a)=-12 my answer, the choices are all wrong

marlon navarro
- 7 years, 2 months ago

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the non zero condition rules out your solution.

Nitin Garg
- 7 years, 2 months ago

a+b+c # 0 and your solution does not imply it.

Mohit Gupta
- 7 years, 2 months ago

there is a condition imposed on this question, i.e. a+b+c $\neq$ 0, where as your choice of -1, 4 and -3 sum up to zero does not satisfy the condition given

Kwong Chin Lam
- 7 years, 2 months ago

LOL :D

Jeremiah Jocson
- 7 years, 2 months ago

sum of these 3 numbers should not be zero

b sathyanarayana
- 7 years, 2 months ago

This does not hold, as this relies on the assumptions that a+b+c=0, which contradicts the statement

Nanayaranaraknas Vahdam
- 7 years, 2 months ago

It is given that a+b+c is not zero...:p

Sanchay Jain
- 6 years, 8 months ago

But in question they have mentioned that a+b+c is not equal to zero

Greeshma Reddy
- 5 years, 7 months ago

There're only a solution for a,b and c is a = b = c = 12^(1/3). All other solutions that sastifies the two conditons do not sastify the non-zero condition. Try to factor a^3+b^3+c^3-3abc (which equal to 0), you will figure out why.

Đức Việt Lê
- 7 years, 2 months ago

Where you got (a+b+c)(a^2+b^2+c^2-ab-bv-ca)?

Steven S
- 7 years, 2 months ago

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(a+b+c)(a^2+b^2+c^2-ab-bc-ca) = a^3 + b^3 + c^3 -3abc.

Niranjan Khanderia
- 7 years, 2 months ago

If we take a=3 b=2 and c=1 That means a^3+b^3+c^3 = 36 and a+b+c is not zero. (a+b)(a+c)(b+c)= 2abc+a^2b+a^2c+b^2c+b^2a+c^2a+c^2b If we put a^2b+a^2c+b^2c+b^2a+c^2a+c^2b = X We ll notice that : (a+b+c)^3 = 3X+ a^3+b^3+c^3+6abc So (a+b)(a+c)(b+c)= 2abc + [(a+b+c)^3-(a^3+b^3+c^3)-6abc] / 3 = 24 + [216-36-72]/3 = 60 Is this solution correct ?!

Zineb Boulaajaj
- 7 years, 2 months ago

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a
*
b
*
c=12 is also a condition.that does not meet your solution.

Sayantan Yadav
- 7 years, 2 months ago

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exactly ri8.. @Sayantan Yadav

pallavi mohini
- 7 years, 2 months ago

Yeeah you're right :) thank uu !!

Zineb Boulaajaj
- 7 years, 2 months ago

may be but not sure as values may nt be taken of own wish..? a^3+b^3c^3=3abc whch means its 3*abc vch is 36..

pallavi mohini
- 7 years, 2 months ago

Agree with u

Reazul Zannat
- 7 years, 2 months ago

abc = 12, but this is not so in your case.

Niranjan Khanderia
- 7 years, 2 months ago

why it couldn't be -12

Satwik Banchhor
- 7 years, 2 months ago

This problem can be solved just using MA >= MG. In this case MA = MG then a = b = c.

Cézar Robert
- 7 years, 2 months ago

Nice solution

Mayank Chaturvedi
- 7 years, 2 months ago

great solution!

Firman Aditama
- 7 years, 2 months ago

Huh?

A Former Brilliant Member
- 7 years, 2 months ago

If a,b,c are real no. and same den abc cannot be equal to 12

Mayank Chawla
- 7 years, 2 months ago

Awesome explanation sir !!!

Soutrik Bandyopadhyay
- 7 years, 2 months ago

A+B+C=0 is not applicable,our question is wrong.

vishal deshmukh
- 7 years, 2 months ago

Nice s I thought the same but u r first to state.

alok kumar
- 6 years, 5 months ago

Since there are no conditions for $a, b, c$ being equal or not, then we can assume that $a = b= c$

Then, $abc = a^3 = 12$ which gives us $a=\sqrt[3]{12}$ . We can verify if this is true by using the first condition. That is, we can check that when $a = \sqrt[3]{12}$ , $a^3 + b^3 + c^3 = 3a^3 = 36$ .

Finally, $(a+b) (b+c) (c+a) = (2a)(2a)(2a) = 8a^3 = 8(\sqrt[3]{12})^3 = 8\cdot 12 = 96$

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There better be some logic behind our assumption !! Otherwise it becomes trial and error method. Which is permitted in many cases.

Niranjan Khanderia
- 7 years, 2 months ago

so, can assume a=b=c in any condition except the one you mentioned above?

Archiet Dev
- 5 years, 6 months ago

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(a^3 + b^3 + c^3) / 3 = abc i.e. AM = GM therefore a=b=c=(12)^(1/3)

there fore ans = 96

Mohit Banerjee
- 7 years, 2 months ago

a^3+b^3+c^3-3abc= (a+b+c)(a^2+b^2+c^2-ab-bc-ca)

now a^3+b^3+c^3= 36 & 3abc=36 By this we can see tht RHS= LHS is zero. It is also given (a+b+c) is not zero so a^2+b^2+c^2-ab-bc-ca=0

(a-b)^2 + (b-c)^2 + (c-a)^2 = 2a^2+2b^2+2c^2-2ab-2bc-2ca=0

only possible when a=b=c

(a+b)(b+c)(c+a)= 2a2b2c=8abc=8*12=96

Anurag Mukhopadhyay
- 7 years, 2 months ago

a^3+b^3+c^3= 3abc since abc=12

By AM-GM Inequality, (a^3+b^3+c^3)/3 ≥ cube root of (a^3)(b^3)(c^3)

(a^3+b^3+c^3)/3 ≥ abc

(a^3+b^3+c^3) ≥ 3abc

Equality holds if and only if a=b=c

Therefore,

3a^3=36

a^3=12

Solving for the given expression: (a+b)(b+c)(c+a)

=(2a)(2a)(2a)

=8a^3

=8(12)

=96

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AM-GM only holds for non-negative values. And we are gives that a,b,c are real numbers, so you cannot aply AM-GM. Try for example a = -5, b = -7 c = 1. You'll see am-gm is not valid in this cases

Juan rodrígez
- 6 years, 9 months ago

*
((a-b)^2+(b-c)^2+(c-a)^2)
RHS=0=LHS
But ATQ,
a+b+c<>0
(a-b)^2+(b-c)^2+(c-a)^2=0
Since square of a number cannot be negative,
a=b=c
(a+b)(b+c)(c+a)=2a2b2c=8abc=8
*
12=96

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a^3+b^3+c^3-3abc= (a+b+c)(a^2+b^2+c^2-ab-bc-ca)

since a^3+b^3+c^3= 36 & 3abc=36 LHS is zero. It is also given (a+b+c) is not zero so a^2+b^2+c^2-ab-bc-ca=0

(a-b)^2 + (b-c)^2 + (c-a)^2 = 2a^2+2b^2+2c^2-2ab-2bc-2ca=0

only possible when a=b=c

(a+b)(b+c)(c+a)= 2a2b2c=8abc=8*12=96

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$a^3+b^3+c^3=36=12+12+12,abc=12\longrightarrow a^3b^3c^3=12\times12\times12$

Just a simple AM-GM and $a=b=c= ^3\sqrt12$ ,SO ANSWER IS $\boxed{96}$

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What made you assume a^3 / b^3 / c^3 = 12?

Alexander Lemere
- 7 years, 2 months ago

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a^3+b^3+c^3-3abc= (a+b+c)(a^2+b^2+c^2-ab-bc-ca)

since a^3+b^3+c^3= 36 & 3abc=36 LHS is zero. It is also given (a+b+c) is not zero so a^2+b^2+c^2-ab-bc-ca=0

(a-b)^2 + (b-c)^2 + (c-a)^2 = 2a^2+2b^2+2c^2-2ab-2bc-2ca=0

only possible when a=b=c

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

2b2c=8abc=8*12=96