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From statement 1, x^3=y^3.
X and y can take positive and negative value both.
If x greater than 0.No conclusion can be drawn from this.
When combining these two we get.
X as positive and y as positive both equal.Hence C.

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Is \( X = Y\) ?

1.\(\sqrt{X^6} = Y^3\)
2.\(X > 0\)

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\(\sqrt{X^6} \)= \( x^{\frac{6}{2}}\) = \(x^3\)
so \(x^3 = y^3\) which implies x = y

Can x be negative ?
Even roots of negative number are not real, so expression inside the square root has to be positive and in the case in hand no matter x is negative or positive \(x^6\) will be positive. so the condition is always met for for negative or positive values of x.

Statement 1 is sufficient IMO, as statement 2 is not a necessary condition.

what is OA?


I went with C when i came across this question on the Test!

If statement 2 was there, it was for a reason.
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Additionally, From Statement 1- we can infer y is always positive; However, x can be negative or positive
so statement 1 is not sufficient

Statement 2 alone is clearly not sufficient

combined
x and y both are positive and will be equal

(C) is correct, I missed considering that y is always positive.


sumitkrocks
hD13
Is \( X = Y\) ?

1.\(\sqrt{X^6} = Y^3\)
2.\(X > 0\)

Posted from my mobile device[/m]

\(\sqrt{X^6} \)= \( x^{\frac{6}{2}}\) = \(x^3\)
so \(x^3 = y^3\) which implies x = y

Can x be negative ?
Even roots of negative number are not real, so expression inside the square root has to be positive and in the case in hand no matter x is negative or positive \(x^6\) will be positive. so the condition is always met for for negative or positive values of x.

Statement 1 is sufficient IMO, as statement 2 is not a necessary condition.

what is OA?
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hD13
Is \(x = y\) ?

(1) \(\sqrt{x^6} = y^3\)
(2) \(x > 0\)

Is x = y?

1)\(\sqrt{x^6} = y^3\)
\(x^{\frac{1}{6}} = y^3\)
Raising both sides to the power of 6:
\(x = y^18\)
Here, x could be 0, y = 0
or x could be 1, y = -1 - Insufficient!
2)\(x > 0\)
Clearly insufficient!
Combining (1) & (2):
x can be 0, y = 0 or x can be 1, y = 1. Either ways, x = y - Sufficient!
IMO, (C)!
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