using identities x-y=6 ; xy=4, find the value of x3-y3?
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Answered by
2
Given x - y = 6 and xy = 4
(x - y)^2 = 36
x^2 + y^2 - 2xy = 36
x^2 + y^2 - 8 = 36
x^2 + y^2 = 44 ---- (1)
x^3 - y^3 = (x-y)(x^2+y^2+xy)
= 6(44 + 4)
= 6(48)
= 288.
(x - y)^2 = 36
x^2 + y^2 - 2xy = 36
x^2 + y^2 - 8 = 36
x^2 + y^2 = 44 ---- (1)
x^3 - y^3 = (x-y)(x^2+y^2+xy)
= 6(44 + 4)
= 6(48)
= 288.
Answered by
0
(x-y)^2= x^2 +y^2- 2xy
(6)^2 = x^2 + y^2 - 2*4
36 +8= x^2 + y^2
44= x^2 + y^2
now
x3-y3= (x-y) (x2+y2 + 2xy)
x3-y3= (6) (44+4)
= 6* (48)
=288
(6)^2 = x^2 + y^2 - 2*4
36 +8= x^2 + y^2
44= x^2 + y^2
now
x3-y3= (x-y) (x2+y2 + 2xy)
x3-y3= (6) (44+4)
= 6* (48)
=288
Anonymous:
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Sorry for interrupting bro. I think x^3-y^3 = (x-y)(x^2+y^2+xy)
ya u r right i m really sorry
No problem bro. U can correct it now
okk thankq by the way
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