coolhabhi wrote:Let α and β be the roots of the equation x^2 - 3x - 3 = 0, then find the value of (β/α^2) + (α/β^2)
a)18
b)3
c)9
d)6

mevicks' solution is AWESOME, but as he stated, this question most likely goes beyond the scope of the GMAT

I just wanted to add a few things about how mevicks reached the conclusion that α + β = 3 and αβ = -3

First recognize that we can't factor the expression x² - 3x - 3
Now, we COULD apply the quadratic formula to find the values of β and α, and then calculate (β/α²) + (α/β²), but the GMAT would never require us to make so many tedious and time-consuming calculations.

Instead, we might make a few observations about the roots of quadratic equations in the form x² - somethingx - something = 0

Some examples
x² - 2x - 15 = 0 is the same as (x - 5)(x + 3) = 0. So the roots are 5 and -3
Notice that 5 + (-3) = 2 and (5)(-3) = 15

x² - 5x - 6 = 0 is the same as (x - 6)(x + 1) = 0. So the roots are 6 and -1
Notice that 6 + (-1) = 5 and (6)(-1) = -6

x² - 4x - 12 = 0 is the same as (x - 6)(x + 2) = 0. So the roots are 6 and -2
Notice that 6 + (-2) = 4 and (6)(-2) = -12

So, if α and β are the roots of x² - 3x - 3 = 0, then we know that α + β = 3 and αβ = -3

We want to evaluate (β/α²) + (α/β²)
Notice that (β/α²) + (α/β²) = (β³/α²β²) + (α³/α²β²) = (β³+α³)/α²β²

Evaluate the denominator: α²β²
Since we already know that αβ = -3, we can see that....
α²β² = (αβ)²
= (-3)²
= 9

Evaluate the numerator: β³ + α³
THIS is where we have problems.
As mevicks pointed out, we need to recognize that β³ + α³ = (β + α)³ - 3βα(β + α)
This is a pretty ingenious observation, one that the GMAT would not require one to make under very severe time constraints. Likewise, the GMAT does not require us to know how to factor a sum of cubes this way [or the other way . . . a³ + b³ = (a + b)(a² - ab + b²)]

For these reasons, I believe this question is out of scope.

Cheers,
Brent