CBSE Class 9 Answered
(x+y)3 = x3 +y3 +3xy(x+y)
(x+y)3 = 5+3xy(x+y)..............(1)
(x2+y2 )(x+y) = 3(x+y)
x3 +y3 +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)
t=2 ==> x+y=2
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