Question : If $2 x-y=2$ and $x y=\frac{3}{2}$, then what is the value of $x^3-\frac{y^3}{8}?$
Option 1: $\frac{9}{2}$
Option 2: $-\frac{5}{4}$
Option 3: $\frac{5}{2}$
Option 4: $\frac{13}{4}$
Correct Answer: $\frac{13}{4}$
Solution :
$2x-y=2$
⇒ $x-\frac{y}{2}=1$ [Dividing both sides by 2]
Also, $xy=\frac{3}{2}$
So, $x^3-\frac{y^3}{8}$
= $(x-\frac{y}{2})^3+3(x)(\frac{y}{2})(x-\frac{y}{2})$
= $(1)^3+\frac{3}{2}(\frac{3}{2})(1)$
= $1+\frac{9}{4}$
= $\frac{13}{4}$
Hence, the correct answer is $\frac{13}{4}$.
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