Question 60 (a)
What should be subtracted from 2x3−3x2y+2xy2+3y3 to get x3−2x2y+3xy2+4y3?
In order to get the solution, we will subtract x3−2x2y+3xy2+4y3 from 2x3−3x2y+2xy2+3y3.
Required expression is
2x3−3x2y+2xy2+3y3−(x3−2x2y+3xy2+4y3)= 2x3−3x2y+2xy2+3y3−x3+2x2y−3xy2−4y3
On combining the like terms.
=2x3−x3−3x2y+2x2y+2xy2−3xy2+3y3−4y3=x3−x2y−xy2−y3
So, if we subtract x3−x2y−xy2−y3 from 2x3−3x2y+2xy2+3y3. then we get x3−2x2y+3xy2+4y3.