if x+y+z=9 and xy+yz+zx=26
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if X + Y + Z equal to 9 and xy + Y Z plus ZX equal to 26 find the value of x cube + y cube + Z cube minus 3 x y z
Use this identity: x³ + y³ + z³ - 3xyz = (x+y+z)(x²+y²+z²-xy-yz-zx)
i try this identity
Answers
Answered by
18
Hey there!
The answer is in the image.
Hope it helps,
Purva
Brainly Community
The answer is in the image.
Hope it helps,
Purva
Brainly Community
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thanx i understand my mistake
Yeah. Sure. You are welcome
Answered by
2
Answer:
Step-by-step explanation:
(x+y+z)^2 = x^2 + y^2 + z^2 + 2 (xy + yz + zx)
(9)^2 = x^2 + y^2 + z^2 + 2 × 26
81 = x^2 + y^2 + z^2 + 52
x^2 + y^2 + z^2 = 81 - 52 = 29
Now, x^3 + y^3 + z^3 - 3xyz
= (x+y+z) [x^2 + y^2 + z^2 + 2 (xy + yz + zx)]
= 9 × 29 + 2(26)
= 9 × 29 + 52
= 9 × 81
= 729
Therefore, x^3 + y^3 + z^3 - 3xyz = 729
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