X+y+z=5 andxy+yz+zx=10 then prove that X cube+y cube+z cube-3xyz=-25
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1
Answer:
(-25)
Step-by-step explanation:
x^3 + y^3 + z^3 - 3xyz =
(x+y+z)(x^2 + y^2 + z^2 - xy - yz - zx)
=5{x^2+y^2+z^2-(xy+yz+zx)}
=5(x^2+y^2+z^2-10)
=5(x^2+y^2+z^2)-50
You know that,
(x+y+z)^2=x^2+y^2+z^2+2xy+2yz+2zx
Now,
x^2+y^2+z^2=(x+y+z)^2-(2xy+2yz+2zx)
x^2+y^2+z^2=5^2-2(xy+yz+zx)
x^2+y^2+z^2=25-2(10)
=5
Substituting x^2+y^2+z^2 by 5,
x^3+y^3+z^3-3xyz
=5(5)-50
=25-50
=(-25)
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