Algebra. Prove that: [tex]\displaystyle pq^2+rs^2-\frac{(pq)^2+(rs)^2+2pqrs}{p+r}=\frac{pr(q-s)^2}{p+r}[/tex]

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[tex]\displaystyle pq^2+rs^2- \frac{(pq)^2+(rs)^2+2pqrs}{p+r} \\= \frac{(p+r)(pq^2+rs^2) - (p^2q^2+r^2s^2+2pqrs))}{p+r} \\= \frac{p^2q^2+r^2s^2+prs^2+prq^2-p^2q^2+r^2s^2-2prqs}{p+r}\\= \frac{prq^2 - 2prqs +prs^2}{p+r} \\= \frac{pr(q^2-qs+s^2)}{p+r}\\= \frac{pr(q-s)^2}{p+r} \\\\\\\star\bigstar\boxed{\boxed{\bold{Terbukti}}}\bigstar \star [/tex]