Question

If x+y+z=1, x^2+y^2+z^2= 2 and x^3+y^3+z^3=3. Then find the value of xyz

Answer 1

x + y + z = 1

x² + y² + z² = 2

x³ + y³ + z³ = 3

________ [GIVEN]

We have to find xyz

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x + y + z = 1

• Squaring on both sides

We know that (a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca

=> (x + y + z)² = x² + y² + z² + 2xy + 2yz + 2zx

=> (1)² = 2 + 2(xy + yz + zx)

=> 1 = 2 + 2(xy + yz + zx)

=> - 1 = 2(xy + yz + zx)

=> xy + yz + zx = $\dfrac{-1}{2}$

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Also..

x³ + y³ + z³ - 3xyz = (x + y + z) (x² + y² + z² - xy - yz - zx)

=> x³ + y³ + z³ - 3xyz = (x + y + z) [x² + y² + z² - (xy + yz + zx)]

Now put the known values in above eq

=> 3 - 3xyz = (1) [(2) - $\dfrac{(-1)}{2}$

=> 3 - 3xyz = 1 × $\dfrac{5}{2}$

=> - 3xyz = $\dfrac{5}{2}$ - 3

=> - 3xyz = $\dfrac{-1}{2}$

=> - xyz = $\dfrac{-1}{6}$

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xyz = $\dfrac{1}{6}$

__________ [ANSWER]