Around the World in 80 Questions (Day 9): Is x^3 - y^3 = 0 ?

statement 1
if xy>0
that does not mean x=y
so not sufficient
statement 2
if x-y =11
that means x is not equals to y
so not sufficient
answer will be e

Is x^3−y^3=0
the question is (x-y)(x^2+xy+y^2)=0?
(1) xy<0
since xy<0 then x is not equal y therefore x^3-y^3 is not equal to 0. Hence it is sufficient. only A and D are left
(2) x−y=11
since x is not equal to y then 2 is sufficient
The answer is D

Q: Is x^3 - y^3 = 0?
Q: Is x^3 = y^3 ?

1) xy < 0
=> One is +ve number and the other is - ve number
=> x^3 - y^3 is not equal to zero ---> SUFFICIENT

2) x - y = 11
=> x = 11 + y
Q: Is (11 + y)^3 = y^3 ?
Cube root both sides => Is 11 + y = y ? No ---> SUFFICIENT

Answer - [D]

My Answer: D, both statements alone are sufficient.

For x^3-y^3 = 0, either x=y or x=y=0.

Statement (1): If xy<0, one of the numbers is positive and the other is negative. So x is not equal to y. Also, it is not possible for x=y=0 because 0 is not less than 0. Hence, sufficient.
Statement (2): If x-y=11, it can be concluded that neither is x=y nor is x=y=0. So, also sufficient.

the given target condition of \(x^3 - y^3 = 0\)
will be possible only when both x & y are same and of same sign i.e. both are +ve or both are -ve

#1
\(xy < 0\)
for this to be true both x & y have to be of opposite sign
the condition of \(x^3 - y^3 = 0\) will not be true
sufficient
#2
\(x - y = 11\)
case 1
x& y are both +ve
14-3 =11
case 2
x & y are of opposite sign
8-(-3)=11
case 3
c & y are both -ve
-3-(-14) = 11
also for x-y = 11
one has to be odd and other number even so therefore answer to target x^3-y^3 will be always no
sufficient
OPTION D

Is x3−y3=0
(1) xy<0
(2) x−y=11

The question is asking whether x3=y3 or x=0 and y=0.
(1) xy<0 means xy= negative number, which shows either x= negative number or y= negative number. sufficient to answer.
(2) x-y=11 means x is greater than y by 11 meaning it is not equal or equal to each other. Sufficient to answer.
Answer is (D)

The Answer is E
As per statement 1- we only find our about xy<0. However we need more info to determine from the factorised form of x3-y3.
i.e (x-y) (x2+y2+xy) Hence insufficient
As per statement 2- We know x−y=11. However we need to know the values of x and y and their signs to answer the question. Hence insufficient.
Combining both, we know that xy<0. However we don't know the signs or values of x and y.
Say, if we take x=-1 and y=1 for xy<0 then through factorising we get x3-y3=0
However, if we take x=1 and y=-1 then we get x3-y3=11. Hence both are insufficient to answer the question.

answer is D.

STATEMENT 1 : xy<0
x and y are of opposite signs. so x^3 and y^3 will be of opposite signs. in such a case x^3 can never be equal to y^3. so x^3 - y^3 is NOT equal to zero. so, this statement is sufficient

STATEMENT 2 : x-y=11
x^3 - y^3 = (x-y)(x^2 + y^2 - xy) = (x-y)*((x-y)^2 + xy) . This can be zero only is xy = -((x-y)^2) i.e., xy = -121 when x-y=11. there exists no such solution. so x^3 - y^3 is NOT equal to zero. so, this statement is sufficient

Therefore each option is sufficient. thus answer is D

Statement 1 is sufficient as according to it either of x or y has to be negative for the product to be less than zero and when such values are put in the equation the solution will be a non zero integer.
Statement 2 is alone not sufficient as we cannot find a specific value for x and y through it using which we can answer the question
Answer A

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Asked: Is \(x^3 - y^3 = 0\) ?

\(x^3 - y^3 = (x-y)(x^2 + y^2 + xy) \)
Is x- y=0; Is x=y?
or
Is x^2 + y^2 + xy = (x+y)^2 - xy = 0?
Is (x+y)^2 = xy

(1) \(xy < 0\)
If x=y; xy = x^2 >=0;
Since xy<0; x is NOT = y
(x+y)^2 = xy < 0 is NOT possible
(x+y)^2 is NOT = xy
SUFFICIENT

(2) \(x - y = 11\)
x-y = 11 is NOT = 0
But (x+y)^2 may or may not = xy
NOT SUFFICIENT

IMO A

\(x^3-y^3=(x-y)(x^2+y^2+xy); x^2+y^2+xy\) is always \(>=0\)

S1) \(xy<0 \) means \(x\) & \(y \) are of opp signs hence term is never \(0\) (Suff)
S2) \(x-y=11\) means, expression is \(11(y^2+11y+121)\).This has a discriminant \(<0\). hence always \(>0\) (suff)

Hence D)

The question asks if x=y

(1) xy<0 -> One must be positive and the other is negative -> x=\=y -> Sufficient

(2) x-y=11 -> x=11+y -> x=\=y -> Sufficient

Answer D

Is x^3−y^3=0
x^3=y^3
x=y

Statement (1) xy<0
means x not equal to y
x & y are of opposite signs
Hence x^3−y^3=0 is not true
Statement 1 is sufficient

(2) x−y=11
x=y+11
x is not equal to y
hence x^3−y^3=0 is not true
Statement 2 is sufficient

Option D is Correct

Excellent question! Kudos to the amazing brain who designed it.
So we need to see if \(x^3 - y^3 = 0\).
That means, (x-y)(x^2+y^2+xy)=0
Hence, either we need x=y or we need x^2+y^2+xy=0
The second case is not possible as you will need imaginary roots to make x^2+y^2+xy=0.
Hence, we need to see of x=y.
(1) xy<0: if x=y, xy will never be less than 0. Hence, sufficient.
(2) x=y+11. Hence, x is not equal to y. Sufficient.

Hence, D is the answer.

Is \(x^3 - y^3 = 0\) ?

(1) \(xy < 0\)

(2) \(x - y = 11\)

Case I: xy <0
what it says is if x >0, y <0 or vice-versa
lets say if x = 2, y = -3
and hence x^3- y^3 is not equal to 0
if x =2 and y = -2
then x^3 -y^3 = 16
so for all the different values the expression x^3 -y^3 is not equal to 0
hence this is sufficient

Case II: x-y =11
x^ - y^3 = (x-y) (x^2+ y^2 +xy)
=(x-y) ((x-y)^2 + 3xy)
= 11 *(121 + 3xy)
again if x and y are both >0 or both <0, then the above expression will be positive and not equal to 0
if x and y are opposite xy<0
in order for the above expression to be 0, 3xy =-121 and x - y =11
i.e 3(11+y) y = -121
3y^2 +33y +121=0
here(33)^2- 4*3*121 = 1089 - 1452 <0
so it will have imaginary

hence there won't be any value of x and y where this expression will be 0
so this is sufficient

D is the answer

Is x^3−y^3=0
(x-y) (x^2 + y^2 + xy) = 0
x - y = 0
x=y


(1) xy<0
If it's true, x is not equal to y
+ + = +
- - = +
Insuff

(2) x−y=11
x = y+11
If it's true, x is not equal to y
Insuff

Combined
Insuff

Ans E

Is x^3−y^3=0
on solving we get is x=y?


(1) xy<0
this says that x and y has opposite signs, so x cannot equal to y, hence sufficient.


(2) x−y=11
difference of x and y =11, so x cannot equal to y, hence sufficient.

Hence my pick will be D, either of (1) or (2) can solve the question.