If x-y=4 and xy=12 find x^3-y^3
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Ravenclaw6405:
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To solve tHis You have to Apply some identities which are...
1. x^3-y^3 = (x-y)(x^2 + y^2 + xy)
2.(x-y)^2 = x^2 y^2 -2xy as....
from 2 we get the value of Xsquare + Ysquare to put in 1.....which is 40 and now put these all the values in 1
ans we get..
Xcube - Ycube = 4×52=208
1. x^3-y^3 = (x-y)(x^2 + y^2 + xy)
2.(x-y)^2 = x^2 y^2 -2xy as....
from 2 we get the value of Xsquare + Ysquare to put in 1.....which is 40 and now put these all the values in 1
ans we get..
Xcube - Ycube = 4×52=208
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