Question

find the value of x³+y³+z³- 3xyz if x²+y²+z²=83.and x+y+z=15​

Answer 1

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Answer 2

Answer:

180

Step-by-step explanation:

We know that,

$(x+y+z)^2=x^2+y^2+z^2+2xy+2yz+2zx\\=>(15)^2=83+2(xy+yz+zx)\\=>225-83 = 2(xy +yz+zx)\\=>142/2=xy+yz+zx\\=>71=xy+yz+zx\\x^3+y^3+z^3-3xyz=(x+y+z)(x^2+y^2+z^2-xy-yz-zx)\\=>x^3+y^3+z^3-3xyz=(15){83-(71)}\\=>x^3+y^3+z^3-3xyz=(15)(12)\\=>x^3+y^3+z^3-3xyz=180$

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