Problem Solving

[GMAT math practice question]

x^3-y^3 = 90 and x-y = 3. What is the value of xy?

A. 3
B. 4
C. 5
D. 6
E. 7

=>

x^3-y^3 = (x-y)(x^2+xy+y^2) = 90
Since x - y = 3, x^2+xy+y^2 = 30
Now, 9 = (x-y)^2 = x^2 - 2xy + y^2 = x^2+xy+y^2 - 3xy = 30 - 3xy.
So, 3xy = 21 and xy = 7.

Therefore, the answer is E.
Answer: E

$$x^3-y^3=90\ ---\ \left(1\right)$$
$$x-y=3\ ---\ \left(2\right)$$
From eqn. (1)
$$x^3-y^3=\left(x-y\right)\left(x^2+xy+y^2\right)\ =90\ \ \ \ \ \ ---\ \left(ii\right)$$
Then from eqn. (2), 3 = x-y
$$Therefore,\ x^3-y^3=3\left(x^2+xy+y^2\right)\ =90\ \ $$
$$\left(x^2+xy+y^2\right)\ =\frac{90}{3}\ =\ 30\ \ \ \ \ -------\left(3\right)$$
$$On\ squaring\ both\ sides\ of\ eqn\ .2,\ we\ have,$$
$$\left(x-y\right)^2=9$$
$$\left(x-y\right)\left(x+y\right)=9$$
$$x^2-xy-xy+y^2=9$$
$$x^2-2xy+y^2=9\ \ \ ---------\left(4\right)$$
Subtract eqn. (4) from eqn. (3) (elimination), we have
3xy = 21
$$xy=7$$
$$Answer\ is\ option\ E$$