If $(3x + 2y)^3 + (3x - 2y)^3 = 3kx(3x^2 +4u^2)$, then the value of k will be :

6

We know that,

a³ + b³ = ( a + b ) ( a² + b² - ab )

Given,

 (3x + 2y)³ + (3x – 2y)³ = 3kx(3x² + 4y²)      ----(A)

We have (3x + 2y)³ + (3x – 2y)³

= {(3x + 2y) + (3x -2y)}{(3x + 2y)² + (3x - 2y)² - (3x + 2y)(3x - 2y)}

= 6x{(9x² + 4y² + 12xy) + (9x2 + 4y2 - 12xy) - (9x² - 4y²)}

= 6x(9x² + 12y²)

= 3 × 6x(3x² + 4y²)      ----(B)

Now if we compare the coefficients of equation(B) with (A), we get

k = 6