if x+y+z=9 and zy+yz+zx=23 then find the value of(x³+y³+z³-3xyz)
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Answered by
68
> (x3 + y3 + z3 - 3xyz)
= (x + y + z) (x2 + y2 + z2 - xy - yz - zx)
= (x + y + z) [(x + y + z)2 - 3(xy + yz + zx)]
= 9 x (81 - 3 x 23)
= (9 x 12)
= 108
= (x + y + z) (x2 + y2 + z2 - xy - yz - zx)
= (x + y + z) [(x + y + z)2 - 3(xy + yz + zx)]
= 9 x (81 - 3 x 23)
= (9 x 12)
= 108
Answered by
26
Answer:=(x3+y3+z3-3xyz).
=(x+y+z) (x2+y2+z2-xy-yz-zx)
=(X+Y+Z) {(X+Y+Z)2-3(XY+YZ+ZX)}
=9×(81-3×23)
=(9×12)
=108
Step-by-step explanation:
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