BREAKING: Target Test Prep releases Brand New 2026 On Demand GMAT prep course

Redeem
TTP OnDemand GMAT Video Course
Featured Course

Target Test Prep GMAT OnDemand

Scott Woodbury-Stewart’s private virtual classroom — 400 hours of master-class video lessons for the GMAT Focus Edition.

  • 715+ score guarantee — highest in the industry (99th percentile)
  • 52 chapters · 1,500+ lessons · 4,000+ practice questions
  • 400 hours of video · 1,500+ instructor-led HD smartboard lessons
  • 300,000+ students accepted to Harvard, Stanford, Wharton, Booth & Sloan & more
  • 24/7 live support + weekly Zoom office hours with GMAT instructors
  • TTP AI Assist — 24/7 AI-powered virtual tutor for instant help
  • 1,200+ flashcards + AI-powered study assistant & daily calendar
  • OnDemand, LiveTeach & GMAT Bootcamp formats available
  • Also: GRE, SAT Math & Executive Assessment courses
  • MBA Admissions Consulting now available
  • 🏆 2025 EdTech Breakthrough Award: Test Prep Solution Provider of the Year
  • 200,000+ students served
  • 5-day free trial — $0 to start, no auto-billing, cancel anytime
★★★★★ 5.0 (559 reviews)
130-pt guarantee
$0 to start then $127/mo
Start Free 5-Day Trial Read 559 Reviews →

x^3+y^3=? 1) x+y=0 2) x^2-y^2=0

Expert replies
Post new topic Post Reply
Previous Topic Next Topic
User avatar
Elite Legendary Member
Posts: 3991
Joined: Fri Jul 24, 2015 2:28 amFollowed by: 37 members
Location: Las Vegas, USA
Thanked: 19 times
by Max@Math Revolution » Tue Apr 26, 2016 5:36 am
x^3+y^3=?

1) x+y=0
2) x^2-y^2=0


* A solution will be posted in two days.
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Source: — Data Sufficiency |
User avatar
Elite Legendary Member
Posts: 3991
Joined: Fri Jul 24, 2015 2:28 amFollowed by: 37 members
Location: Las Vegas, USA
Thanked: 19 times

by Max@Math Revolution » Fri Apr 29, 2016 7:10 pm
Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution.

x^3+y^3=?

1) x+y=0
2) x^2-y^2=0


-> In the original condition, there are 2 variables(x,y), which should match with the number of equations.
For 1) 1 equation, for 2) 1 equation, which is likely to make C the answer.
When 1) & 2), you can only substitute 1)x=-y.
Then, x3^+y^3=(-y)^3+y^3=-y^3+y^3=0 is derived, which is unique and sufficient.
Hence, the answer is A.
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Post new topic Post Reply
• Page 1 of 1