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, $$x^3 + y^3 + z^3 – 2 = 12 \times (96 – 24)$$
$$\Rightarrow \frac{ x^3 + y^3 + z^3}{4} = 216.5$$