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