Question

x+y+z=3, x²+y²+z²=101, then find √x³+y³+z³-3xyz
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Answer 1

Answer:

x+y+z=3, x²+y²+z²=101, then find √x³+y³+z³-3xyz

Answer 2

The value of √(x³ + y³ + z³ - 3xyz) is 21.

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• The expression x³ + y³ + z³ - 3xyz can be expanded as :

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

• We have the value of  ( x + y + z ) and ( x² + y² + z² ) stated in the question, but the value of xy - yz - xz is unknown.

• Therefore, to calculate the value of xy - yz - xz, we need to apply a formula that contains both ( x + y + z ) and ( x² + y² + z² ), so that with the help of their values, the value of xy - yz - xz can be determined.

• The required formula is stated below :

( x + y + z )² = x² + y² + z² + 2 ( xy + yz + xz ) -(ii)

• Putting x + y + z = 3, and x² + y² + z² = 101 in the above expansion, we get,

3² = 101 + 2 ( xy + yz + xz )

=> 9 = 101 + 2 ( xy + yz + xz )

=> 2 ( xy + yz + xz ) = 9 - 101

=> 2 ( xy + yz + xz ) = - 92

=> ( xy + yz + xz ) = - 92 / 2

=> ( xy + yz + xz ) = - 46

=> - ( - xy - yz - xz ) = - 46

=> - xy - yz - xz = 46

• Now, substituting all the values on the right hand side of equation (i), we get,

x³ + y³ + z³ - 3xyz = 3 × ( 101 + 46 )

=> x³ + y³ + z³ - 3xyz = 3 × 147

=> x³ + y³ + z³ - 3xyz = 441

=> √(x³ + y³ + z³ - 3xyz) = √441

=> √(x³ + y³ + z³ - 3xyz) = 21 (Answer)

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