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Notice that \(a^3 + b^3 + c^3 - 3abc = (a + b + c)(a^2 + b^2 + c^2 - (ab + bc +ca))\)

 

The only missing info we need is the value of \(a^2 + b^2 + c^2\).

 

Now, because \((a + b + c)^2 = a^2 + b^2 + c^2 + 2(ab + bc + ca)\), plugging in values gives \(a^2 + b^2 + c^2 = 11^2 - 2(25) = 71\)

 

You can plug in the values on your own.

so like what maxwong said we have a^2+b^2+c^2=71 and a^3 + b^3 + c^3 - 3abc = (a + b + c)(a^2 + b^2 + c^2 - (ab + bc +ca))  plug that in to get 11*(71-25) and that equals 506.

 

 

 

 

 

_{^{Credts: Maxwong for hints}}