If $$x+y+z=12$$ and $$x^2+y^2+z^2=70$$, find the value of $$x^3+y^3+z^3−3xyz$$ .

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Uniteasy Admin · Accepted Answer

Consider the identity of square of trinomial
$$(x+y+z)^2 = x^2+y^2+z^2+2xy+2yz+2zx$$n
Plug in the given conditions
 $$x+y+z=12\x^2+y^2+z^2=70$$
The value of $$xy+yz+zx$$ is obtained
 $$xy+yz+zx = (12^2-70)/2 = 37$$
Now using the identity
$$x^3+y^3+z^3−3xyz = (x+y+z)(x^2+y^2+z^2−xy−yz−zx)$$
$$= 12\cdot (70-37) =396$$

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