Question

12x cube +9y cube plz solve it fast.......

Answer 1

Given :

= > 12x^3 + 9y^3

= > 3(4x^3 + 3y^3)

$= \ \textgreater \ 3( \sqrt[3]{4}x)^3 + ( \sqrt[3]{3} y)^3$

We know that a^3 + b^3 = (a + b)(a^2 - ab + b^2)

$= \ \textgreater \ ( \sqrt[3]{4} x + \sqrt[3]{3} y)(( \sqrt[3]{4} )^2x^2 - \sqrt[3]{3} \sqrt[3]{4} xy + ( \sqrt[3]{3} )^2y^2)$

$= \ \textgreater \ 3( \sqrt[3]{4} x + \sqrt[3]{3} y)( 4^{ \frac{2}{3} }x^2 - \sqrt[3]{12}xy + 3^ \frac{2}{3} y^2)$



Hope this helps!

Answer 2

Hi,

Please see the attached file!



Thanks