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Asked by studdybuddy | 29 Sep, 2008, 07:34: PM Expert Answer

(x+y)3 = x3 +y3 +3xy(x+y)

(x+y)3 = 5+3xy(x+y)..............(1)

now

x2+y2=3

(x2+y2 )(x+y) = 3(x+y)

x3 +y3 +xy(x+y) = 3(x+y)

5+xy(x+y)= 3(x+y)

xy(x+y)= 3(x+y) -5.....(2)

now put value of xy(x+y) in(1) from (2)

(x+y)3 = 5+3( 3(x+y) -5 )

(x+y)3 -9(x+y) +10=0

let x+y =t

after solving this qubic equation

we get (t-2)(t2+2t+5) = 0

since t2+2t+5= (t+1)2+4>0 ( always positive)

so t-2=0

t=2 ==> x+y=2

Answered by | 14 Dec, 2008, 02:20: AM
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