Question

if $3x+2y=12$ and $xy=6$ then find the value of ${27x}^3+{8y}^3$

Answer 1

It is given that $3x+2y=12$ and $xy=6$.

We find the value of $27x^3+8y^3$ as shown below:

$27x^3+8y^3=(3x{)}^3+(2y{)}^3$

$=(3x+2y)\left[\right(3x{)}^2-(3x\times 2y)+\left(2y{)}^2\right](\because a³+b³=(a+b\left)\right(a²-ab+b²\left)\right)$

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

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

$=12\left[\right(3x{)}^2-36+\left(2y{)}^2\right]$

$=12\left[\right(3x{)}^2+(2y{)}^2+(2\times 3x\times 2y)-(2\times 3x\times 2y)-36]$

$=12\left[\right(3x+2y{)}^2-12xy-36]$

$=12\left[\right(12{)}^2-(12\times 6)-36]$

$=12(144-72-36)$

$=12(144-108)$

$=12\times 36$

$=432$

Hence, $27x^3+8y^3=432$.