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Value of abc expression = ?

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by GMATGuruNY » Fri Nov 25, 2016 6:48 am
The problem should indicate that abc ≠ 0.
coolhabhi wrote:If a + b + c = 0 then the value of

[(a2 + b2 - ab)/ab] + [(b2 + c2 - bc)/bc] + [(c2 + a2 - ca)/ca] is

a) -6
b) -3
c) -1
d) 0
e) 1
Let a=1, b=1 and c=-2, with the result that a+b+c = 1+1-2 = 0.
Substituting these values into the given expression, we get:

[(1 + 1 - 1)/1] + [(1 + 4 + 2)/-2] + [(4 + 1 + 2)/-2] = 1 - 7/2 - 7/2 = -6.

The correct answer is A.
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by coolhabhi » Fri Nov 25, 2016 6:59 am
GMATGuruNY wrote:The problem should indicate that abc ≠ 0.
coolhabhi wrote:If a + b + c = 0 then the value of

[(a2 + b2 - ab)/ab] + [(b2 + c2 - bc)/bc] + [(c2 + a2 - ca)/ca] is

a) -6
b) -3
c) -1
d) 0
e) 1
Let a=1, b=1 and c=-2, with the result that a+b+c = 1+1-2 = 0.
Substituting these values into the given expression, we get:

[(1 + 1 - 1)/1] + [(1 + 4 + 2)/-2] + [(4 + 1 + 2)/-2] = 1 - 7/2 - 7/2 = -6.

The correct answer is A.
Thanks Mitch. You really are GMATGuru. I have been banging my head against the wall to crack this problem. I could however solve it only after 5 mins of strenuous brain work. You, sire have solved it in just 20 seconds. Thank You.
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by Matt@VeritasPrep » Fri Nov 25, 2016 2:27 pm
Algebraically, we're asked to find:

((a - b)² + ab)/ab + ((b - c)² + bc)/bc + ((c - a)² + ca)/ca

Assuming that a, b, and c are all nonzero, this becomes

a/b + b/a - 1 + b/c + c/b - 1 + c/a + a/c - 1

Now let's use a trick:

(a/b + c/b) = (a + c)/b

Since (a + b + c) = 0, we know (a + c) = -b, so this is just -b/b, or -1.

Doing some color-coding, we can use this trick three times:

a/b + b/a - 1 + b/c + c/b - 1 + c/a + a/c - 1

(a + c)/b - 1 + (b + c)/a - 1 + (a + b)/c - 1

-b/b - 1 + -a/a - 1 + -c/c - 1

-1 + -1 + -1 + -1 + -1 + -1

-6
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by Matt@VeritasPrep » Fri Nov 25, 2016 2:29 pm
I should point out that under strict timed conditions, my approach is much worse than Mitch's ... but this is obviously NOT a GMAT question, and since it looks more like a proof question from a challenging algebra class, I think it's a lot more fun to see why it works.
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