Question

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

Answer 1

Given that

$2x+3y=12$ …… (1)

$xy=6$ ……. (2)

On taking cube both sides in equation (1), we get

${(2x+3y)}^3={\left(12\right)}^3$

$8x^3+27y^3+3\times 2x\times 3y(2x+3y)=1728$

$8x^3+27y^3+18xy(2x+3y)=1728$

From equation (1) and (2),

$8x^3+27y^3+18\times 6\times 12=1728$

$8x^3+27y^3+1296=1728$

$8x^3+27y^3=1728-1296$

$8x^3+27y^3=432$

Hence, this is the answer.