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y^3 - t^3 or x^3 - y^3 is the same if you are more comfortable

(x-y) * (x^2 - y^2) -----> (x - y) * (x - y) * (x + y)

1) x - y = 3 -----> 3 * 3 * (x + y) ----> we do no have the value od (x + Y)

2) we have 21 BUT not (x - y)

1) + 2) we have 3 * 21 ----> Sufficient

;)
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carcass
y^3 - t^3 or x^3 - y^3 is the same if you are more comfortable

(x-y) * (x^2 - y^2) -----> (x - y) * (x - y) * (x + y)

1) x - y = 3 -----> 3 * 3 * (x + y) ----> we do no have the value od (x + Y)

2) we have 21 BUT not (x - y)

1) + 2) we have 3 * 21 ----> Sufficient

;)

Hi Carcass,

You are right that it does not make any difference between y,t & x,y.

But you made a mistake in the formula \(x^3-y^3\) is not equal to \((x-y)\) * \((x^2 - y^2)\)
Instead \((x-y)^3\) is equal to \(x^3-y^3\) + \(3xy(x-y)\)

I Hope it helps.
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carcass
y^3 - t^3 or x^3 - y^3 is the same if you are more comfortable

(x-y) * (x^2 - y^2) -----> (x - y) * (x - y) * (x + y)

1) x - y = 3 -----> 3 * 3 * (x + y) ----> we do no have the value od (x + Y)

2) we have 21 BUT not (x - y)

1) + 2) we have 3 * 21 ----> Sufficient

;)

Hi Carcass,

You are right that it does not make any difference between y,t & x,y.

But you made a mistake in the formula \(x^3-y^3\) is not equal to \((x-y)\) * \((x^2 - y^2)\)
Instead \((x-y)^3\) is equal to \(x^3-y^3\) + \(3xy(x-y)\)

I Hope it helps.
:shocked :wall ...........oh lord, after a long day of work the gmat is even worse.

Some Mod could say how to attack this question properly AND in the faster manner
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What is the value of y^3 - t^3?

(1) y - t = 3. If \(y=3\) and \(t=0\), we have one value for \(y^3 - t^3\) and for \(y=4\) and \(t=1\) another. Not sufficient.

(2) y^2 - t^2 = 21. The same here. Not sufficient.

(1)+(2) From (2) we have \((y-t)(y+t)=21\). Since from (1) \(y - t = 3\), then \(3(y+t)=21\) --> \(y+t=7\). So we have two distinct linear equation with two unknowns: \(y+t=7\) and \(y - t = 3\), which means that we can solve for them and get the value of \(y^3 - t^3\). Sufficient.

Answer: C.

Hope it's clear.
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