Question

plz help me out... ...

Answer 1

22) from the identity
if a+b+c =0 then
a³+b³+c³= 3abc
therefore 20³+30³-50³=3 abc
= 3 x 20 x30 x - 50
here a= 20

b= 30
c= - 50

Answer 2

22...)
30³+20³-50³
=3³*10³+2³*10³+5³*10³
=(3³+2³+5³)10³
=(27+8+125)1000
=170000

23...)
x+y+z=1.......................(i)
xy+yz+xz=(-1)..............(ii)
xyz= -1.........................(iii)

we know that.
x³ + y³ + z³– 3xyz = (x + y + z) (x²+ y² + z²– xy – yz – zx).
x³+y³+z³ = (x+y+z)[x²+y²+z²-(xy+yz+zx)]+3xyz
x³+y³+z³= 1 [x²+y²+z² +1] -3
x³+y³+z³= x²+y²+z²-2