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## Comment on

Tricky Simplification## I used a different approach

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?

## Be careful.

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

Cheers,

Brent

## 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?

## It's possible, but unlikely.

It's possible, but unlikely.

## I choose x=4 and y=2. But

## Hi Kaori,

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.

Cheers,

Brent

## Why didn't you simplify the

The x+y in the denominator cancels that in the numerator leaving x^2?

## Be careful, x³ + y³ does not

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?

Cheers,

Brent

## If I choose x=2 and y=1, then

## You're right. With those

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?

Cheers,

Brent

## Hi Brent, instead of testing

## Can you show me how your

Can you show me how your approach would look?

## Multiply N & D by conjugate

(x³+y³)(x-y)/(x-y)(x+y)

{(x^4-x³y+y³x-y^4)}/(x²-y²)

{(x^4-y^4)-xy(x²-y²)}/(x²-y²)

{(x²-y²)(x²+y²)-xy(x²-y²)}/(x²-y²)

(x²-y²)(x²+y²)/(x²-y²)-xy(x²-y²)/(x²-y²)

(x²+y²)-xy

## Very nice!!

Very nice!!

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

Cheers,

Brent

## I think as a thumb rule,

## That's a great idea, but in

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: https://www.gmatprepnow.com/module/gmat-powers-and-roots/video/1044

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.

Cheers.

Brent

## hi brent, just for my

## There's not really a useful

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.

Cheers,

Brent

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