(x - y)^3(x−y)3
Solution
Well you can use many methods to simplify like:
Using Pascal Triangle which give be 1, 3, 3, 11,3,3,1 as the expansion..
You can simplify (x - y)^3(x−y)3 to either (x - y) (x - y) (x - y) or (x - y)^2 (x - y)(x−y)(x−y)(x−y)or(x−y)2(x−y)
But using those two will result in same answer which will be in this format ->→ 1, 3, 3, 11,3,3,1
Hence rArr⇒ (x - y)^3 = (x - y) (x - y) (x - y)(x−y)3=(x−y)(x−y)(x−y)
(x - y) (x - y) (x - y)(x−y)(x−y)(x−y)
(x - y) [(x - y) (x - y)](x−y)[(x−y)(x−y)]
(x - y) [x^2 - xy - xy + y^2](x−y)[x2−xy−xy+y2]
This lead to the point using difference of two cubes as:
(x - y) [x^2 - 2xy + y^2](x−y)[x2−2xy+y2]
x [x^2 - 2xy + y^2] - y [x^2 - 2xy + y^2]x[x2−2xy+y2]−y[x2−2xy+y2]
x^3 - 2x^2y + xy^2 - x^2y + 2xy^2 - y^3x3−2x2y+xy2−x2y+2xy2−y3
Collect like terms
color(red)(x^3 - y^3) color(blue)(- 2x^2y - x^2y) color(green)(+ xy^2 + 2xy^2) x3−y3−2x2y−x2y+xy2+2xy2
:. x^3 - y^3 - 3x^2y + 3xy^2 -> Answer
If it is the cube of a binomial, it will be in this format rArr color(red)(1)x^3 - color(red)(3)x^2y + color(red)(3)xy^2 - color(red)(1)y^3
