Question : What will be the value of $x^{3}+y^{3}+z^{3}-3xyz$, when $x+y+z=9$ and $x^{2}+y^{2}+z^{2}=31?$
Option 1: 27
Option 2: 3
Option 3: 54
Option 4: 9
Correct Answer: 54
Solution : Give: $x+y+z=9$ ......(1) $x^{2}+y^{2}+z^{2}=31$ ......(2) As $(a+b+c)^2 = a^2 + b^2 + c^2 + 2(ab+bc+ca)$ Squaring (1) on both sides, we have, $(x+y+z)^2 = 81$ ⇒ $x^2 + y^2 + z^2 +2(xy+yz+zx) = 81$ ⇒ $31+2(xy+yz+zx) = 81$ ⇒ $xy+yz+zx = 25$ Now, $x^{3}+y^{3}+z^{3}-3xyz$ = $ (x+y+z)(x^2 + y^2 + z^2 - xy - yz- zx)$ = $(9)(31-25) = 9\times 6 = 54$ Hence, the correct answer is 54.
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