Question

x+y=3,xy=2 so find x³-y³=?

Answer 1

x+y=3,xy=2
so x-y = root over of (x+y)^2-4xy = root over of 3^2-4×2=root over of 1=1
x-y=1
then
x^3-y^3=(x-y)(x^2+xy+y^2)=1[(x^2+2xy+y^2)-xy] = 1[(x+y)^2-xy]
=3^2-2=9-2=7
so the ans is 7


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Answer 2

Answer:

7

Explanation:

As x+y=3

let x & y be 2 & 1 respectively

so, 2+1 = 3

   2*1=2

as this proves right that x = 2 & y = 1

lets replace x³-y³ with the above numbers

which will give

                      =  2³ - 1³

                      =  (2*2*2) - (1*1*1)

                      =  8 - 1

                      = 7

∴  x³-y³

=  2³ - 1³

=7