Suppose x−y=1. Find the value of x4−xy3−x3y−3x2y+3xy2+y4.
Answer:
1
- Given x−y=1
We need to find the value of x4−xy3−x3y−3x2y+3xy2+y4. - x4−xy3−x3y−3x2y+3xy2+y4=(x4−x3y)−(xy3−y4)−(3x2y−3xy2)[Rewriting the expression]=x3(x−y)−y3(x−y)−3xy(x−y)=(x−y)[x3−y3−3xy]=(x−y)[(x−y)(x2+xy+y2)−3xy][Using a3−b3=(a−b)(a2+ab+b2)]=(x2+xy+y2)−3xy[Given x−y=1]=x2−2xy+y2=(x−y)2[Using (a−b)2=a2−2ab+b2]=1[Given x−y=1]
- Hence, the value of the given expression is 1.