factorise (2y+x)2 (y-2x) +(2x+y)2 (2x-y)

(2x-y) [-4y2 -4xy-x2+4x2+y2+4xy)

=(2x-y)(3x2 -3y2)

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

  • 9

The answer is (2x+y) (2x+y) (4x).

  • 1

can you please re check the question. i am not sure but i think in the first term it should be (2y - x)

  • -2

it is a good question. took a lot of time. i think your question is wrong. if in the first term you would replace (y - 2x) with (2y - x) then the solution would be-

[(2y+x)2(2y-x)] + [(2x+y)2(2x-y)]

[(4y2+x2+4xy)(2y-x)] + [(4x2+y2+4xy)(2x-y)]

[8y3+2x2y+8xy2-4xy2-x3-4x2y] + [8x3+2xy2+8x2y-4x2y-y3-4xy2]

[8y3-x3-2x2y+4xy2] + [8x3-y3-2xy2+4x2y]

8y3 - y3 + 8x3 - x3 + 2xy2 + 2x2y

7y3 +7x3 + 2xy(x+y)

7(x3 + y3) + 2xy(x+y)

(7)(x + y)(x2 + y2 - xy) + 2xy(x + y)

(7)(x+y)(x2 + y2 - xy + 2xy)

(7)(x + y)(x2 + y2 + xy)

  • 2
What are you looking for?