Question

If $2x-3y=10$ and $xy=16$ ; find the value of $8x^3-27y^3$

• A. 1
• B. 2
• C. 2, x
• D. No common factors exist

Answer 1

The correct option is C

2, x

Given that $2x-3y=10$
Cubing on both sides we get
$(2x-3y{)}^3={10}^3$
$8x^3+27y^3-3\left(2x\right)\left(3y\right)(2x-3y)=1000$

Given that $xy=16$
$8x^3+27y^3-18\left(16\right)\left(10\right)=1000$
$8x^3+27y^3-2880=1000$
$8x^3+27y^3=3880$