Answer: Identity:x3 +y3 +z3 -3xyz = (x+y+z)( x2 +y2 +z2 -xy-yz-zx) Since x + y + z = 1, xy + yz + zx = -1and xyz= -1, Putting values, we get: So Now (x+y+z)2 = x2 +y2 +z2 +2xy +2yz +2zx 12 = x2 +y2 +z2 +2(-1) x2 +y2 +z2 = 3 Put in eq (i) x3 +y3 +z3 =3-2 = 1 Answered by | 4th Jun, 2014, 03:23: PM Concept Videos Practice Test All Questions Ask Doubt if x+1/x =7 then find x cube + 1/x cubw (a+b+c)^3 13×13×13+7×7×7/13×13-13×7+7×7 factorise 8x*3+y*3+27z*3+18xyz if x+1/x=5,then find value of x^3+1/x^3 The valuesof 249square -248square is 729X3-512-y3 Factorise : (a+b+c)³-a³-b³-c3 Reply
Answer:
Identity:x3 +y3 +z3 -3xyz = (x+y+z)( x2 +y2 +z2 -xy-yz-zx)
Since x + y + z = 1, xy + yz + zx = -1and xyz= -1,
Putting values, we get:
So
Now
(x+y+z)2 = x2 +y2 +z2 +2xy +2yz +2zx
12 = x2 +y2 +z2 +2(-1)
x2 +y2 +z2 = 3
Put in eq (i)
x3 +y3 +z3 =3-2 = 1
Answered by | 4th Jun, 2014, 03:23: PM
Concept Videos
Practice Test
All Questions Ask Doubt
if x+1/x =7 then find x cube + 1/x cubw
(a+b+c)^3
13×13×13+7×7×7/13×13-13×7+7×7
factorise 8x*3+y*3+27z*3+18xyz
if x+1/x=5,then find value of x^3+1/x^3
The valuesof 249square -248square is
729X3-512-y3
Factorise : (a+b+c)³-a³-b³-c3