Michelle Mendoza

2022-07-10

Show that $\frac{\left(b+c{\right)}^{2}}{3bc}+\frac{\left(c+a{\right)}^{2}}{3ac}+\frac{\left(a+b{\right)}^{2}}{3ab}=1$
If ${a}^{3}+{b}^{3}+{c}^{3}=3abc$ and $a+b+c=0$ show that $\frac{\left(b+c{\right)}^{2}}{3bc}+\frac{\left(c+a{\right)}^{2}}{3ac}+\frac{\left(a+b{\right)}^{2}}{3ab}=1$

SweallySnicles3

Expert

$a+b+c=0\phantom{\rule{thickmathspace}{0ex}}⟺\phantom{\rule{thickmathspace}{0ex}}b+c=-a\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}\left(b+c{\right)}^{2}={a}^{2}$
$\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}\frac{\left(b+c{\right)}^{2}}{3bc}=\frac{{a}^{3}}{3abc}$
Actually, $b+c=-a\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}\left(b+c{\right)}^{3}=\left(-a{\right)}^{3}$
$\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}-{a}^{3}={b}^{3}+{c}^{3}+3bc\left(b+c\right)={b}^{3}+{c}^{3}+3bc\left(-a\right)\phantom{\rule{thickmathspace}{0ex}}⟺\phantom{\rule{thickmathspace}{0ex}}\sum {a}^{3}=3abc$

ziphumulegn

Expert

$\frac{\left(b+c{\right)}^{2}}{3bc}+\frac{\left(c+a{\right)}^{2}}{3ac}+\frac{\left(a+{b}^{2}\right)}{3ab}\phantom{\rule{0ex}{0ex}}=\frac{a\left(b+c{\right)}^{2}}{3abc}+\frac{b\left(c+a{\right)}^{2}}{3abc}+\frac{c\left(a+{b}^{2}\right)}{3abc}\phantom{\rule{0ex}{0ex}}=\frac{{a}^{3}}{3abc}+\frac{{b}^{3}}{3abc}+\frac{{c}^{3}}{3abc}\phantom{\rule{0ex}{0ex}}=\frac{{a}^{3}+{b}^{3}+{c}^{3}}{3abc}\phantom{\rule{0ex}{0ex}}=\frac{3abc}{3abc}=1$