Basic Algebra test part 4

First, let's simplify things a little. Let $p = 6bcx, q = 3acy, r = 2abz$ . By the given equation, we get $p + q + r = 0$ . Also let $\displaystyle\alpha = \frac{q-r}{p}, \beta = \frac{r-p}{q}, \gamma = \frac{p-q}{r}$ . Now let's prove two Lemmas:

Lemma 1: $\alpha+\beta + \gamma = -\alpha\beta\gamma$

Proof: $\alpha + \beta + \gamma = \frac{q-r}{p} + \frac{r-p}{q} + \frac{p-q}{r} = \frac{qr(q-r)+pr(r-p)+pq(p-q)}{pqr}$ $= \frac{1}{pqr}\left(\frac{(r-q)^3 +q^3 - r^3}{3} + \frac{(p-r)^3 +r^3 - p^3}{3} + \frac{(q-p)^3 +p^3 - q^3}{3}\right)$ $=\frac{(r-q)^3+(p-r)^3+(q-p)^3}{3pqr}$ Now there is a theorem which states that if $X+Y+Z=0$ , then $X^3+Y^3+Z^3 = 3XYZ$ . Therefore, we can simplify further: $\alpha + \beta + \gamma = \frac{(r-q)(p-r)(q-p)}{pqr} = -\alpha\beta\gamma. \boxed{}$ Lemma 2: $\alpha\beta+\beta\gamma+\gamma\alpha = -9$ .

Proof: $\alpha\beta + \beta\gamma +\gamma\alpha = \frac{(q-r)(r-p)}{pq} + \frac{(r-p)(p-q)}{qr} + \frac{(p-q)(q-r)}{rp}$ $= \frac{-r^2+r(p+q)-pq}{pq} + \frac{-p^2+p(q+r)-qr}{qr} + \frac{-q^2+q(p+r)-rp}{rp}$ $=-3-\frac{2r^2}{pq}-\frac{2p^2}{qr}-\frac{2q^2}{rp} = -3 -2\left(\frac{p^3+q^3+r^3}{pqr}\right)$ Now we use the same theorem from Lemma 1, and we get $\alpha\beta + \beta\gamma +\gamma\alpha = -3-2*3 = -9. \boxed{}$

So, let's finally prove what we need to prove. The expression that we need to simplify becomes: $\left(\frac{q-r}{p}+\frac{r-p}{q}+\frac{p-q}{r}\right)\left(\frac{p}{q-r}+\frac{q}{r-p} + \frac{r}{p-q}\right)$ $= (\alpha+\beta+\gamma)\left(\frac{1}{\alpha}+\frac{1}{\beta}+\frac{1}{\gamma}\right) = \frac{(\alpha+\beta+\gamma)(\alpha\beta+\beta\gamma+\gamma\alpha)}{\alpha\beta\gamma}$ By Lemmas 1 and 2, this simplifies to $\boxed{9}$ .

So simple, $\frac{3acy-2bza}{6bcx}=(\frac{y}{2b}+\frac{z}{3c})\frac{a}{x}$ Because $\frac{x}{a}+\frac{y}{2b}+\frac{z}{3c}=0$ , so $\frac{y}{2b}+\frac{z}{3c}=-\frac{x}{a}$ Lastly $(\frac{y}{2b}+\frac{z}{3c})\frac{a}{x}=(-\frac{a}{x})\frac{x}{a}=-1$ Do the same thing to another then get 9

Let (3acy-2bza)/(6bcx) = m

(2abz-6bcx)/(3acy) = n

(6bcx-3acy)/(2abz) = p

So

(m+n+p)x(1/m+1/n+1/p) =

3+(m/n+n/m)+(m/p+p/m)+(n/p+p/n)

AM-GM

3+2+2+2= 9

CMIIW