Question

\(x + y + z = 12 \) , \(x^2 + y^2 + z^2 = 96 \) and \( \frac{1}{x} + \frac{1}{y} + \frac{1}{z} = 36\). Find the value of \( \frac{x^3 + y^3 + z^3}{4} \).

Solution:

we know , \((a + b + c)^2 = a^2 + b^2 + c^2 + 2(ab + bc + ac)\)
then, \((x + y + z)^2 = 144 = 96 + 2(xy + yz +xz)\)
\(\Rightarrow 24 = xy + yz + xz\)

Given , \( \frac{1}{x} + \frac{1}{y} + \frac{1}{z} = 36 = \frac{xy + yz + xy}{xyz} \)
\(\Rightarrow 24 = 36xyz\)
\(\Rightarrow xyz = \frac{2}{3} \).
\(\Rightarrow 3xyz = 2\)

we know \(a^3 + b^3 + c^3 – 3abc = (a + b + c) (a^2 + b^2 + c^2 – ab – bc – ac)\)
then, \(x^3 + y^3 + z^3 – 2 = 12 \times (96 – 24)\)
\( \Rightarrow \frac{ x^3 + y^3 + z^3}{4} = 216.5 \)