Pls help me solving this
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Squaring both the sides,
(x+y+z)²=(9)²
x²+y²+z²+2(xy+yz+zx)=81
xy+yz+zx= 23(given)
x²+y²+z²+2(23)=81
x²+y²+z²+46=81
x²+y²+z²=81-46
x²+y²+z²= 35
Now,
x³+y³+z³-3xyz=(x+y+z)(x²+y²+z²-xy-yz-zx)
Now put all the value on Formula,
x³+y³+z³-3xyz=(x+y+z){x²+y²+z²-(xy+yz+zx)}
x³+y³+z³-3xyz=(9){35-(23)}
x³+y³+z³-3xyz=(9)(35-23)
x³+y³+z³-3xyz=9x12
x³+y³+z³-3xyz= 108
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