Question

Find the value of
(i) $8x^3+y^3-3xy+125,\text{if}2x+y=-5$
(ii) $x^3-27y^3+27xy+27,ifx=3y-3$

Answer 1

$i).$ Value of $8x^3+y^3-3xy+125$

Given : $2x+y=-5$

This can be written as $2x+y+5=0$

Let $2x=9,$ $y=b,$ $c=5$

$\because a+b+c=0$

Therefore, $a^3+b^3+c^3=3abc$

$\Rightarrow {\left(2x\right)}^3+{\left(y\right)}^3+{\left(5\right)}^3=3\left(2x\right)\left(y\right)\left(5\right)$

$\Rightarrow 8x^3+y^3+125=30xy$

$\therefore 8x^3+y^3-3xy+125=30xy-3xy=27xy$

$ii).x^3-27y^3+27xy+27$

Given : $x=3y-3$

This can be written as $x-3y+3=0$

Let $x=a,-3y=b,3=c$

$\because a+b+c=0$

$\Rightarrow a^3+b^3+c^3=3abc$

$\Rightarrow x^3+{(-3y)}^3+3^3=3\left(x\right)(-3y)\left(3\right)$

$\Rightarrow x^3-27x^3+27=-27xy$

$\therefore x^3-27y^3+27xy+27=27xy-27xy=0$

Hence, solved.