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A fast way to solve this question? I solved it, but it took me more than 2 min :s.

What is the value of \(y^3 - t^3\)? (1) \(y - t = 3\) (2) \(y^2 - t^2 = 21\)

I spent a lot of time in solving the equations with the possible values of y and t.

Source: gmathacks.com

(1) - (y-t) = 3 ----> Y & t can be anything -Insufficient (2) (y+t)(y-t) = 3 x 7 ----> (y+t) can be 3 or 7 -----> i.e. y will be 5 for sure but t can be 2 or -2 ----> Insufficient (1+2) (y+t)= 7 because (y-t) = 3----> y = 5 & t=2--->Sufficient

Hope it helps. _________________

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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. _________________

If you like my Question/Explanation or the contribution, Kindly appreciate by pressing KUDOS. Kudos always maximizes GMATCLUB worth-Game Theory

If you have any question regarding my post, kindly pm me or else I won't be able to reply

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.

...........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 _________________

(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.

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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