( a + b ) ( b + c ) ( c + a )
a , b and c are real numbers such that a + b + c = 0 , a 3 + b 3 + c 3 = 3 6 , a b c = 1 2 . What is the value of ( a + b ) ( b + c ) ( c + a ) ?
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8 solutions
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
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the non zero condition rules out your solution.
a+b+c # 0 and your solution does not imply it.
there is a condition imposed on this question, i.e. a+b+c = 0, where as your choice of -1, 4 and -3 sum up to zero does not satisfy the condition given
LOL :D
sum of these 3 numbers should not be zero
This does not hold, as this relies on the assumptions that a+b+c=0, which contradicts the statement
It is given that a+b+c is not zero...:p
But in question they have mentioned that a+b+c is not equal to zero
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.
Where you got (a+b+c)(a^2+b^2+c^2-ab-bv-ca)?
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(a+b+c)(a^2+b^2+c^2-ab-bc-ca) = a^3 + b^3 + c^3 -3abc.
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 ?!
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a b c=12 is also a condition.that does not meet your solution.
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exactly ri8.. @Sayantan Yadav
Yeeah you're right :) thank uu !!
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..
Agree with u
abc = 12, but this is not so in your case.
why it couldn't be -12
This problem can be solved just using MA >= MG. In this case MA = MG then a = b = c.
Nice solution
great solution!
Huh?
If a,b,c are real no. and same den abc cannot be equal to 12
Awesome explanation sir !!!
A+B+C=0 is not applicable,our question is wrong.
Nice s I thought the same but u r first to state.
Since there are no conditions for a , b , c being equal or not, then we can assume that a = b = c
Then, a b c = a 3 = 1 2 which gives us a = 3 1 2 . We can verify if this is true by using the first condition. That is, we can check that when a = 3 1 2 , a 3 + b 3 + c 3 = 3 a 3 = 3 6 .
Finally, ( a + b ) ( b + c ) ( c + a ) = ( 2 a ) ( 2 a ) ( 2 a ) = 8 a 3 = 8 ( 3 1 2 ) 3 = 8 ⋅ 1 2 = 9 6
There better be some logic behind our assumption !! Otherwise it becomes trial and error method. Which is permitted in many cases.
so, can assume a=b=c in any condition except the one you mentioned above?
Assume a=b=c, 3a3=36(from a3 + b3 + c3 =36), axaxa=12 (from abc=12) a3=12 hence assumption is true. put b=c=a and find the answer, i.e, 96
(a^3 + b^3 + c^3) / 3 = abc i.e. AM = GM therefore a=b=c=(12)^(1/3)
there fore ans = 96
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
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
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
a^3+b^3+c^3-3abc =((a+b+c)/2) ((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
We can easily conclude that a 3 + b 3 + c 3 = 3 a b c , so from the AM-GM inequality we know that a = b = c . Then a = b = c = 3 1 2 and ( a + b ) ( b + c ) ( c + a ) = 9 6 .
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
a 3 + b 3 + c 3 = 3 6 = 1 2 + 1 2 + 1 2 , a b c = 1 2 ⟶ a 3 b 3 c 3 = 1 2 × 1 2 × 1 2
Just a simple AM-GM and a = b = c = 3 1 2 ,SO ANSWER IS 9 6
What made you assume a^3 / b^3 / c^3 = 12?
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 2b 2c=8abc=8*12=96