Question

If x+y+z = 8 and xy+yz+zx =20, find the value of x³+y³+z³-3xyz
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Answer 1

Answer:

given,,,

x+y+z=8

xy+yz+zx=20

so . squaring both sides

(x+y+z)^2 = 8^2

x^2 + y^2 +z^2 +2xy +2yz+2zx = 64

x^2 +y^2+z^2 +2(xy+yz+zx)=64

x^2+y^2+z^2+2× 20=64

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

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

x^2 +y^2+z^2= 24

now..

x^3 +y^3+z^3-3xyz= (x+y+z)(x^2+y^2+z^2-xy-yz-zx)

x^3+y^3+z^3-3xyz= 8(24-(xy+yz+zx)

x^3+y^3+z^3-3xyz= 8(24-20)

= 8×4

= 32

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