Alg2.1.28

Show that $(a+b+c)^{3}=27abc\,$ if $(a^{\frac{1}{3}}+b^{\frac{1}{3}}+c^{\frac{1}{3}})=0\,$

Given that $\sqrt[3]{a}+\sqrt[3]{b}+\sqrt[3]{c}=0\,$

$\sqrt[3]{a}+\sqrt[3]{b}=-\sqrt[3]{c}\,$

Cubing on both sides of the equation,we get

$(\sqrt[3]{a}+\sqrt[3]{b})^3=(-\sqrt[3]{c})^3\,$

Expanding the lefthand side expression

$a+3\sqrt[3]{a^2}\sqrt[3]{b}+3\sqrt[3]{a}\sqrt[3]{b^2}+b=-c\,$

Keeping a,b,c on the left hand side we get

$a+b+c=-3\sqrt[3]{a^2}\sqrt[3]{b}-3\sqrt[3]{a}\sqrt[3]{b^2}\,$

Taking out the common term on the right hand side,it looks like

$a+b+c=-3\sqrt[3]{a}\sqrt[3]{b}(\sqrt[3]{a}+\sqrt[3]{b}\,$

Substituting the value

$\sqrt[3]{a}+\sqrt[3]{b}=-\sqrt[3]{c}\,$

in the equation

$a+b+c=-3\sqrt[3]{a}\sqrt[3]{b}(-\sqrt[3]{c})\,$

Cubing on both sides

we get

$(a+b+c)^{3}=(3\sqrt[3]{a}\sqrt[3]{b}\sqrt[3]{c})^{3}\,$

$(a+b+c)^{3}=27abc\,$