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arjtryarjtry
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Should be C, as nothing can be cancelled out from the fraction, and we need to know the relationship b/w (a& b) AND (b&c) to determine what will be the sign of the result
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arjtryarjtry
hmm.. wrong...



I got it..
it should be E then
we can say >0 but question is asking about >=0??
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x2suresh
arjtryarjtry
hmm.. wrong...



I got it..
it should be E then
we can say >0 but question is asking about >=0??

sorry guys, slight mistake in the problem.. it should be cube, not square in the second term of Dr i.e (b-c)^3
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arjtryarjtry
IF a,b,c are distinct non zero nos,

is (a+b)^2 * (b-c) >=0 ?
(a-b)^3 * (b-c)^3

1) a>b
2)b>c


1/(a-b)>=0?

1) a>b
a-b>0 1/(A-B)>0

A is suffcient.

2) not sufficient .

A is the answer.
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IF a,b,c are distinct non zero nos,

is (a+b)^2 * (b-c) >=0 ?
(a-b)^3 * (b-c)^3

1) a>b
2)b>c



Problem can be written as :

(a+b)^2>=0 ?
(a-b)^3 * (b-c)^2

S1. we do not know whether a and b are + or -ve.
numerator will always be +ve
(b-c)^2 will be +ve even though we do not know the value of c

NOW
even if a is +ve and b is -ve we will still get (a-b)^3 as positive ex: (2- (-5))^3. Even if a is -ve and a>b we will get a +ve value for (a-b)^3 ex: (-1 -(-2)). And even if both are =ve we will ge a +ve value for (a-b)^3. SUFF. BCE are out adn AD stays

S2. Tells you about b and c only and not about the relationship b/w a and b

(a+b)^2_ will always be +ve and so will (b-c)^2
However, (a-b)^3 can be +ve or -ve
ex. (2-3)^3 = -ve
(3 -2)^ 3 = +ve
hence MAYBE

Ans A
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arjtryarjtry
IF a,b,c are distinct non zero nos,

is (a+b)^2 * (b-c) >=0 ?
(a-b)^3 * (b-c)^3

1) a>b
2)b>c

the equation above depends on sign of a-b only since odd power of a-b hence (1) is SUFFI
(2) is not

IMO A
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spriya
arjtryarjtry
IF a,b,c are distinct non zero nos,

is (a+b)^2 * (b-c) >=0 ?
(a-b)^3 * (b-c)^3

1) a>b
2)b>c

the equation above depends on sign of a-b only since odd power of a-b hence (1) is SUFFI
(2) is not

IMO A
oops I misread the question again .. it depends on a-b as well as b-c
hence both (1) and (2) are important to solve
hence IMO C
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I think it should be C

You also need to ensure that a <> b and b <> c. The question statement only states that they all are non zero.

1) Ensures that a-b is +ve, but says nothing about b<>c. The possible options are > 0 and undefined (if b=c)
2) makes sure that b-c <> 0. Now we know that there is only one option >0
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bhushangiri
I think it should be C

You also need to ensure that a <> b and b <> c. The question statement only states that they all are non zero.

1) Ensures that a-b is +ve, but says nothing about b<>c. The possible options are > 0 and undefined (if b=c)
2) makes sure that b-c <> 0. Now we know that there is only one option >0

Question says.. distinct non zero numbers
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arjtryarjtry
IF a,b,c are distinct non zero nos,

is (a+b)^2 * (b-c) >=0 ?
(a-b)^3 * (b-c)^3

1) a>b
2)b>c

(a+b)^2 must be >=0
(b-c)/[(b-c)^3] = 1/[(b-c)^2] must be >0

So only the term (a-b)^3 will determine whether the fraction > or < = 0

(1) sufficient:
a > b
then (a-b)^3 > 0
then the fraction > 0

(2) NOT sufficient
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x2suresh
bhushangiri
I think it should be C

You also need to ensure that a b and b c. The question statement only states that they all are non zero.

1) Ensures that a-b is +ve, but says nothing about bc. The possible options are > 0 and undefined (if b=c)
2) makes sure that b-c 0. Now we know that there is only one option >0

Question says.. distinct non zero numbers

that was a good one...



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