Question

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

Collected in the board: The сube of sum

Steven Zheng posted 2 weeks ago

Answer

Consider the identity of square of trinomial

(x+y+z)^2 = x^2+y^2+z^2+2xy+2yz+2zx
(1)

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

Steven Zheng posted 2 weeks ago

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