Question

if x+y+z=9 and xy+yz+zx=23 the value of x^3+y^3+z^3-3xyz is plzzz fast

Answer 1

Here's your answer !!

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We know that,
$( {x}^{3} + {y}^{3} + {z}^{3} - 3xyz) = (x + y + z) \\ ( {x}^{2} + {y}^{2} + {z}^{2} - xy - yz - zx)$

So,

$= > (x + y + z) {(x + y + z)}^{2} - 3(xy + yz + zx)...(1)$

$= > given = x + y + z = 9.....(2) \\ xy + yz + zx = 23....(3)$


By putting value of (2) and (3) in equation (1), we get :-

$= > 9 \times {(9)}^{2} - 3(23) \\ = > 9 \times( 81 - 69) \\ = > 9 \times 12 \\ = > 108$

Hence,

$= > ( {x}^{3} + {y}^{3} + {z}^{3} - 3xyz) = 108$

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Hope it helps you !! :)

Answer 2

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since, x + y + z = 9
and, xy + yz + zx = 23

so, firstly, squaring both side ,

( x + y + z )² = (9)²

=> x² + y² + z² + 2(xy + yz + zx) = 81 ....(using identity)

=> x² + y² + z² + 2(23) = 81

=> x² + y² + z² + 46 = 81

=> x² + y² + z² = 81 - 46

=> x² + y² + z² = 35 ......(i)

Now, by using identity,

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

( substituting the value of x² + y² + z² , from (i) )

x³ + y³ + z³ - 3xyz = (9) [35 - (xy + yz + zx )]

x³ + y³ + z³ - 3xyz = 9 ( 35-23 )

x³ + y³ + z³ - 3xyz = 9 × 12

x³ + y³ + z³ - 3xyz = 108 .....ANSWER
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HOPE IT WAS HELPFUL ✌️!