If x+y+z=9 and xy+yz+zx = 23, the value of (x3+y3+z3−3xyz)=?
(a) 108
(b) 207
(c) 669
(d) 729
Givenx+y+z=9and xy+yz+zx=23
We knowx3+y3+z3−3xyz=(x+y+z)(x2+y2+z2−xy−yz−zx)
But (x+y+z)2=x2+y2+z2+2xy+2yz+2zxx2+y2+z2=(x+y+z)2−2xy−2yz−2zx
So,x3+y3+z3−3xyz=(x+y+z)(x2+y2+z2−xy−yz−zx)=(x+y+z)((x+y+z)2−3xy−3yz−3zx)=(x+y+z)((x+y+z)2−3(xy+yz+zx)=9(92−3×23)=9(81−69)=9×12=108
(a) 108