Question

If $2x+3y=12$ and $xy=6$, find the value of $8x^3+27y^3$.

Answer 1

$2x+3y=12$ $xy=6$

$8x^3+27y^3$

$=(2x{)}^3+(3y{)}^3$

$=a^3+b^3=(a+b)(a^2+b^2-ab)$

$8x^3+27y^3=(2x+3y)\left(\right(2x{)}^2+(3y{)}^2-(2x\left)\right(3y\left)\right)$

$=\left(12\right)\left[\right(2+3y{)}^2-12xy-6xy]$

$=\left(12\right)\left[\right(12{)}^2-18\left(6\right)]$

$=12[144-108]$

$=12\times 36$

$=432$.