2023/010) Given $2x = 3y-5$ find the value of $8x^3-27y^3+90xy +125$

 We have 

$2x = 3y-5$

Cubing both sides as in the required condition we have $8x^3$ and $27y^3$

We get $8x^3 = (3y-5)^3 = (3y)^3) - 5^3 -3(3y)5(3y-5)$ (using $(a -b)^3 = a^3- b^3 -3ab(a-b)$

or  $8x^3 = (3y-5)^3 = (3y)^3 - 5^3 -3(3y)*5 * 2x$ as $3y-5 = 2x$

or $8x^3 = 27y^3- 125 - 90xy$

or $8x^3-27y^3 + 90xy +125 = 0$