Question: Tricky Simplification

Comment on Tricky Simplification

I used a different approach to arrive at C.

X^3+Y^3 (X+Y)^3
-------- = --------
X+Y (X+Y)^1

Deducted denominator from numerator since they now have same base get
(X+Y)^2 (which is one of the Classic quadratics)

i.e x^2+2xy+y^2

Can you please explain why this is wrong?
gmat-admin's picture

Be careful.
x³ + y³ does not equal (x + y)³

We can confirm this by taking (x + y)³ and expanding it to see whether it equals x³ + y³

(x + y)³ = (x + y)(x + y)(x + y)
= (x + y)(x² + 2xy + y²)
= x³ + 2x²y + y²x + yx² + 2xy² + y³
= x³ + 3x²y + 3y²x + y³


Hi Brent,

As the sum of cubes is out of GMAT scope as you mentioned, is it still possible to have such questions in the GMAT considering the alternative solution that could be used to solve it?

gmat-admin's picture

It's possible, but highly unlikely.

I choose x=4 and y=2. But with these assumed values, answers B and E yield correct results. Am I doing something wrong?
gmat-admin's picture

Hi Kaori,

You've done everything right so far.

If it turns out that your chosen pair of values (x = 4 and y = 2) don't eliminate 4 of the 5 answer choices, then you must select a different pair of values and test the remaining answer choices.


Why didn't you simplify the question as (x^3 + y^3)/(x + y) = (x^2)(x+y)/(x+y)?
The x+y in the denominator cancels that in the numerator leaving x^2?
gmat-admin's picture

Be careful, x³ + y³ does not equal (x²)(x + y)

The great thing about factoring, is that we can always check our factoring by expanding the resulting expression.

Take: (x²)(x + y)
Expand to get: x³ + yx² (which is not the same as x³ + y³)

Does that help?


If I choose x=2 and y=1, then choice B seems to work as well.
gmat-admin's picture

You're right. With those values, you'd find that B and E both work.
At that point, you'd have to test another pair of values to eliminate one of those two possible options.

Here's a similar example:

Which of the following is equivalent to 2x + 3x for all values of x?
A) 5x
B) x + 16
C) 10x
D) x + 1
E) x - 100

ASIDE: Of course, we already know that 2x + 3x simplifies to be 5x, but this is meant to illustrate a point.

Let's test a value of x.

How about x = 4
When x = 4, then 2x + 3x = 2(4) + 3(4) = 8 + 12 = 20

Now we'll plug x = 4 into each answer choice to see which one also equals 20

We get:
A) 5x = 5(4) = 20 WORKS
B) x + 16 = 4 + 16 = 20 WORKS
C) 10x = 10(4) = 40 no good
D) x + 1 = 4 + 1 = 5 no good
E) x - 100 = 4 - 100 = -96 no good

We're left with A and B
Now let's test another value of x.

How about x = 11
When x = 11, then 2x + 3x = 2(11) + 3(11) = 22 + 33 = 55

Now we'll plug x = 11 into the two remaining answer choices to see which one also equals 55

We get:
A) 5x = 5(11) = 55 WORKS
B) x + 16 = 11 + 16 = 27 no good

Answer: A

Does that help?


Hi Brent, instead of testing options, how about we just multiply the Num and Den by the conjugate of x+y. Solving it gives the solution in 2 -3 steps.
gmat-admin's picture

Can you show me how your approach would look?

Multiply N & D by conjugate (x-y)
gmat-admin's picture

Very nice!!
The hard part is knowing, in advance, that multiplying the top and bottom by (x - y) will eliminate the denominator.


I think as a thumb rule, whenever we need to modify or get rid of the denominator, the conjugate method should work. Similar to your earlier lessons where you rationalised the denominator with root terms. Eg: _/2+_/3 and we took the conjugate to solve further. But this may not work as smoothly in everycase.
gmat-admin's picture

That's a great idea, but in most cases, that strategy won't eliminate the denominator.

When "fixing" the denominator of a fraction that has a square root in the denominator, multiplying the numerator and denominator by the denominator's conjugate will just get rid of the square root(s) in the denominator. In some cases, the denominator will cancel out, but in must cases, we're still left with some kind of denominator. More here:

The same applies to algebraic fractions.

For example, if we take (x + 5)/(x + y) and multiply top and bottom by (x - y), we get: (x² - xy + 5x + 5y)/(x² - y²)

So, we're still left with a fraction.

That said, multiplying the numerator and denominator by the denominator's conjugate may still be helpful in some cases.


hi brent, just for my understanding, what would be the factoring technique for sums of cubes?
gmat-admin's picture

There's not really a useful technique (within the scope of the GMAT) that we can use to factor a sum of cubes (or to factor a difference of cubes).

Instead, we'd need to memorize the following:

(a³ + b³) = (a + b)(a² - ab + b²)
(a³ - b³) = (a - b)(a² + ab + b²)

That said, these factorizations are beyond the scope of the GMAT.


I used the conjugate of the denominator to cancel it out but got X² -Y² as an answer.

(X³ + Y³)(X - Y)/(X + Y)(X - Y)
(X⁴ - X³Y + Y³X - Y⁴/X² - Y²

And I think this is where I made a mistake. I thought -X³Y + Y³X would cancel each other out leaving me with:

(X⁴ - Y⁴)/(X² - Y²) and then using A^b/A^c = A^(b-c) I found X²-Y²

So my question is, why can't we rewrite -X³Y as -XY³? Is -X³Y = -X³*Y^1 and so as the bases are different you can't sum them up?

gmat-admin's picture

I should start by noting that solving this question algebraically is beyond the scope of the GMAT.
That said, you bring up a good question.

The answer is that the algebraic terms X³Y and XY³ aren't "like terms," which means they can't be combined in the same way that we can add 5k and 3k to get 8k.
In order for two terms to be "like terms" the variable parts must be identical.
So, for example, the following terms are like terms: 4xy³z, 7xy³z, -3xy³z and xy³z
Likewise, these terms are also like terms: 5w²x, 11w²x and -w²x
Since -X³Y and XY³ aren't like terms, we can't combine them.

Here's the video on combining like terms:


exactly what i did, but I'm abit afraid of using 1 because it is too simple and the person who made this question could have this mislead us.
gmat-admin's picture

If the expression in the question is quite complicated, testing a simple value is useful to help eliminate some of the answer choices, at which point you can test a different number with the remaining answer choices.

Hi Brent, so factoring doesn't work with cube root right as below as we can't get back original equation after factoring out? unless its
(a³ + b³) = (a + b)(a² - ab + b²)

(x^3 + y^3)/(x + y) = (x + y)(x^2 + y^2)/(x + y)
= (x + y)(x + y)
= x^2 + 2xy + y^2
gmat-admin's picture

There are a couple of problems with your solution.
For example: (x^3 + y^3) does not equal (x + y)(x^2 + y^2)
To understand why, use FOIL to expand (x + y)(x^2 + y^2).
When you do this you will get x^3 + xy^2 + x^2y + y^3 (not x^3 + y^3)

Also, even IF you were able to get (x + y)(x^2 + y^2)/(x + y)
This expression does not simplify to be (x + y)(x + y)
In actuality, we take: (x + y)(x^2 + y^2)/(x + y)
And cancel the (x + y)'s to get: x^2 + y^2

Thanks Brent. Understand that it only works up to the last two stages.

In actuality, we take: (x + y)(x^2 + y^2)/(x + y)
And cancel the (x + y)'s to get: x^2 + y^2
gmat-admin's picture

That's correct.

(x + y)(x^2 + y^2)/(x + y) = [(x + y)/(x + y)](x^2 + y^2)
= [1](x^2 + y^2)
= x^2 + y^2

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