Question: Tricky Simplification

Comment on Tricky Simplification

I understand your method, but I think that one can find the answer directly if one takes into account that: x^3 + y^3 = (x+y)(x^2 - xy + y^2)


In Nepal, we are taught the sum of cubes formula at school level together with other algebraic expressions. So, I directly used the formula for finding answer to this.

greenlight-admin's picture

I think the sum of cubes factorization is taught in most algebra classes, but most students promptly forget it after they graduate :-)

greenlight-admin's picture

Yes, I mention at the very beginning that we COULD use Sum of Cubes factoring (as you have noted). However, this is beyond the scope of the GRE.

The point of this practice question is to show that, even if we don't know how to factor sums of cubes (which most students don't know how to do), we can still find the correct answer.

ASIDE: I taught high school math for 7 years, and I still don't remember the Sum of Cubes formula :-)

When plugging in values we can use different numbers for x and y right? Because I plugged in x=2 and y=3 to solve the equation and get answer choice E.
greenlight-admin's picture

That works too!

ASIDE: When some students see two different variables (e.g., x and y), they assume that those variables cannot have the same value (e.g., x = 3 and y = 3). This, however, is false. Two different variables CAN have the same value.

I tried to plug in values other than what you used.. I tried to plug (X=2 and Y=3) but found all the options are wrong.. why?
greenlight-admin's picture

It works for the values x = 2 and y = 3

Given: (x³ + y³)/(x + y)
Plug in x = 2 and y = 3
We get: (2³ + 3³)/(2 + 3) = (8 + 27)/(5)
= 35/5
= 7

Now plug x = 2 and y = 3 into all 5 answer choices to see which one evaluates to be 7
E) x² - xy + y²
We get: 2² - (2)(3) + 3² = 4 - 6 + 9
= 7
A match.

This does not seem like a practical strategy for the GRE. This seems like it would eat up way too much time. After all, only evaluating with one set of numbers does not leave me very confident that it would work for other tricky numbers like using 0, or -1. Testing these other numbers then, would take up too much time.
greenlight-admin's picture

The main purpose of this video question is to demonstrate what it means to say one algebraic expression is equal to another expression.

For example, we know that 3x + 7x = 10x, but what does it MEAN?

It means that, FOR ANY VALUE OF X, the expression 3x + 7x will ALWAYS evaluate to have the same value of 10x.

For example, if x = 4, then 3x + 7x = 3(4) + 7(4) = 40
Likewise, 10x = 10(4) = 40

This property (3x + 7x = 10x) will hold true for ALL values of x.

For the video question above, we need only test values until we eliminate 4 of the 5 answer choices.

For example, when x = 1 and y = 1, the original expression evaluates to be 1.
When we plug x = 1 and y = 1 into answer choice A, the expression evaluates to be 0.
As such, we know that x-y does NOT equal the original expression

In the solution, we found that by testing x = 1 and y = 1, we're able to eliminate A, B, C and D.
This means there's no need to test any other values.

Yes, the process could take time, but if you don't have any other means of finding the correct answer, this strategy will work.

For example, what if you were asked to simplify (x² - y²)/(x + y)

With a bit of factoring, we get:
(x² - y²)/(x + y) = (x + y)(x - y)/(x + y) = x - y

However, if you didn't know how to factor the numerator, you could always test values of x and y.

I hope that helps.


First we have a formula (a+b)^2,

Similarly there is (a+b)^3 = a^3 + b^3 +3ab(a+b)

So using that formula a^3+b^3 = (a+b)^3 - 3ab(a+b)

x^3+y^3 can be written as (x+y)^3-3xy(x+y)

So in x^3+y^3 / x+y -->

(x+y)^3 -3xy(x+y)/ x+y ==> (x+y)^3 /x+y - 3xy(x+y)/x+y

==> (x+y)^2*(x+y)/(x+y) - 3xy(x+y)/(x+y)

==> (x+y)^2 - 3xy

==> X^2 + y^2 +2xy -3xy

==> finally, x^2+y^2-xy is the answer.

How about this approach Brent?
greenlight-admin's picture

That's a great approach - nice work!!

Here is my approach: I multiple both numerator & denominator by (x-y) and by solving num & den we can derive the right answer
greenlight-admin's picture

I'm not sure that will work (unless you know already know how to factor the expression x³ + y³)

Given: (x³ + y³)/(x + y)
Multiply top and bottom by (x-y) to get:(x - y)(x³ + y³)/(x + y)(x - y)
Expand and simplify to get: (x⁴ - x³y + xy³ - y⁴)/(x² - y²)
Where do you go from here?

After expanding you can group terms this way, (x^4-y^4)/(x^2-y^2) + xy(x^2-y^2)/(x^2-y^2), simplifying the numerators will give us: x^2 + y^2 + xy
greenlight-admin's picture

We have:
(x⁴ - x³y + xy³ - y⁴)/(x² - y²) = (x⁴ - y⁴)/(x² - y²) + (xy³ - x³y)/(x² - y²)
= (x² + y²)(x² - y²)/(x² - y²) + xy(y² - x²)/(x² - y²)
= (x² + y²) - xy(x² - y²)/(x² - y²)
= (x² + y²) - xy
= x² - xy + y²
= E

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