Question

If x+y+z = 8 and xy+yz+zx =20, find the value of ----
${x}^{3} + {y}^{3} + {z}^{3} - 3xyz$


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Answer 1

Answer:

x+y+z=8

xy+yz+zx=20

(x+y+z)^2=x^2+y^2+z^2+2(xy+yz+zx)

=> (8)^2=x^2+y^2+x^2+2(20)

=> 64=x^2+y^2+z^2+40

=> x^2+y^2+z^2=64-40=24

x^3+y^3+z^3-3xyz

=(x+y+z)(x^2+y^2+z^2-xy-yz-zx)

=8{24-(xy+yz+zx)}

=8(24-20)

=8×4

=32

Answer 2

Answer:

Step-by-step explanation:

Given that,

x + y + z = 8 and xy + yz + zx = 20

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

( 8 )² = ( x² + y² + z² ) + 2 ( 20 )

64 = ( x² + y² + z² ) + 40

64 - 40 = (x²+ y² + z²)

x² + y² + z² = 24

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

= ( 8 ) ( 24 - 20 )

= ( 8 ) ( 4 )

= 32 is the answer.