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Given that x + y + z = 9 and xy + yz + zx = 23.
(x + y + z) = 9
On squaring both sides, we get
(x + y + z)^2 = (9)^2
x^2 + y^2 + z^2 + 2xy + 2yz + 2zx = 81
x^2 + y^2 + z^2 + 2(xy + yz + zx) = 81
x^2 + y^2 + z^2 + 2(23) = 81
x^2 + y^2 + z^2 + 46 = 81
x^2 + y^2 + z^2 = 81 - 46
x^2 + y^2 + z^2 = 35 ----- (1).
Then x^3 + y^3 + z^3 - 3xyz = (x + y + z)(x^2 + y^2 + z^2 - xy - yz - zx)
= (9)(35 - (xy + yz + zx))
= 9(35 - 23)
= 9(12)
= 108.
Hope this helps!
(x + y + z) = 9
On squaring both sides, we get
(x + y + z)^2 = (9)^2
x^2 + y^2 + z^2 + 2xy + 2yz + 2zx = 81
x^2 + y^2 + z^2 + 2(xy + yz + zx) = 81
x^2 + y^2 + z^2 + 2(23) = 81
x^2 + y^2 + z^2 + 46 = 81
x^2 + y^2 + z^2 = 81 - 46
x^2 + y^2 + z^2 = 35 ----- (1).
Then x^3 + y^3 + z^3 - 3xyz = (x + y + z)(x^2 + y^2 + z^2 - xy - yz - zx)
= (9)(35 - (xy + yz + zx))
= 9(35 - 23)
= 9(12)
= 108.
Hope this helps!
SanskritiTripathi11:
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Answered by
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Answer:
sorry bro don't know...........how to solve it........
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