Question

If 8x^3 + 27y^3 = 730 and 3x^2y + 3xy^2 = 15 then evaluate 2x + 3y
Pls help me

Answer 1

Given that, "$\\8x^3 + 27y^3 = 730$ and $3x^2y + 3xy^2 = 15$"

Using this two equations evaluate $2x+3y$:

Using Cube formula for a+b,

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

Plug in a=2x \,\, \, and\, \, \, b=3y in the formula,

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

$=8x^3+27y^3+36x^2y+54xy^2$

We know that the value of $8x^3 + 27y^3 = 730$

Plug in this value in the above equation,

$(2x+3y)^3=730+36x^2y+54xy^2$

Now take 18 as common for the second and third term in the right hand side,

$(2x+3y)^3=730+18(2x^2y+3xy^2)$

Now plug in the value 2x^2y+3xy^2\: \: as \: \: 15,

$(2x+3y)^3=730+18*15$

$(2x+3y)^3=730+270$

$(2x+3y)^3=1000$

Taking Cube root on both sides, we get

$2x+3y=10$