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10 Feb 2021, 07:53
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How to solve this?
a+b+c=0
\(\frac{(a+b)^{11}+ (b+c)^{11} + (a+c)^{11}}{a + a^{11} + b + b^{11} + c + c^{11}}\) = ?
a+b+c=0
\(\frac{(a+b)^{11}+ (b+c)^{11} + (a+c)^{11}}{a + a^{11} + b + b^{11} + c + c^{11}}\) = ?
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10 Feb 2021, 08:01
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Substitute these in the numerator:
a + b = -c
a + c = -b
b + c = -a
The numerator becomes:
(-c)^11 + (-a)^11 + (-b)^11
= - (c^11 + a^11 + b^11)
= - (a^11 + b^11 + c^11)
Denominator equals:
(a+b+c) + (a^11 + b^11 + c^11)
= 0 + (a^11 + b^11 + c^11)
= (a^11 + b^11 + c^11)
Thus, the ratio = -1
Answer = -1
Posted from my mobile device
a + b = -c
a + c = -b
b + c = -a
The numerator becomes:
(-c)^11 + (-a)^11 + (-b)^11
= - (c^11 + a^11 + b^11)
= - (a^11 + b^11 + c^11)
Denominator equals:
(a+b+c) + (a^11 + b^11 + c^11)
= 0 + (a^11 + b^11 + c^11)
= (a^11 + b^11 + c^11)
Thus, the ratio = -1
Answer = -1
Posted from my mobile device
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Schools: Harvard '24 Judge '23 LBS '24 McCombs '24 Mendoza Tepper '24 Ross '24 Cox Marshall '24 Fisher Kelley '24 Jones McDonough '24 Owen '24 Terry BU Simon '24 Katz GWU Mason Babson Gabelli Foster '24 Goizueta '24 Scheller Stern '23 CBS '23 Booth '23 Haas '25 Wharton '25 Stanford '26 Johnson'23 Fuqua '23 Kellogg '23 Kenan-Flagler'23 INSEAD Aug 22 intake Olin '23 Said '23 Sloan '23 Carey Arizona "22 Yale '23 UC Davis Madison '23 Anderson '23 Carroll Carlson '23 Darden '23
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10 Feb 2021, 08:18
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sujoykrdatta
sujoykrdatta
Hello,
Hope you are well in this pandemic.
I did not see this type of question in official gmat books! Is it gmat related question? Can you confirm, please? Thanks__
