# Question-Answer

## If x+y+z=12 and x^2+y^2+z^2=96 and {1}/{x}+{1}/{y}+{1}/{z}=36. Find the value of {x^3+y^3+z^3}/{4}.

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 …