Find the value of x^{3}+y^{3}+z^{3}-3xyz,when | vedclass

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(B) We are given the identity: $x^{3}+y^{3}+z^{3}-3xyz = (x+y+z)(x^{2}+y^{2}+z^{2}-xy-yz-zx)$.First,we find the value of $x^{2}+y^{2}+z^{2}$ using the

Vedclass · Accepted Answer

$352$

Vedclass · Answer

(B) We are given the identity: $x^{3}+y^{3}+z^{3}-3xyz = (x+y+z)(x^{2}+y^{2}+z^{2}-xy-yz-zx)$.First,we find the value of $x^{2}+y^{2}+z^{2}$ using the identity $(x+y+z)^{2} = x^{2}+y^{2}+z^{2}+2(xy+yz+zx)$.Substituting the given values: $16^{2} = x^{2}+y^{2}+z^{2} + 2(78)$.$256 = x^{2}+y^{2}+z^{2} + 156$.$x^{2}+y^{2}+z^{2} = 256 - 156 = 100$.Now,substitute these values into the original identity:$x^{3}+y^{3}+z^{3}-3xyz = (16)(100 - 78)$.$x^{3}+y^{3}+z^{3}-3xyz = 16(22) = 352$.

Find the value of $x^{3}+y^{3}+z^{3}-3xyz$,when $x+y+z=16$ and $xy+yz+zx=78$.

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