MHB Prove Identity: $b_1x^3=b_2y^3=b_3z^3$ & $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=1$

[Albert1]

(1):
$b_1x^3=b_2y^3=b_3z^3$
(2):
$\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=1$
prove:
$\sqrt[3]{b_1x^2+b_2y^2+b_3z^2}=\sqrt[3] {b_1}+\sqrt[3] {b_2} + \sqrt[3] {b_3}$

 

[mente oscura]

Re: prove the indentity

(1):
$b_1x^3=b_2y^3=b_3z^3$
(2):
$\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=1$
prove:
$\sqrt[3]{b_1x^2+b_2y^2+b_3z^2}=\sqrt[3] {b_1}+\sqrt[3] {b_2} + \sqrt[3] {b_3}$

Hello.

Spoiler

\sqrt[3]{b_1x^2+b_2y^2+b_3z^2}=

=\sqrt[3]{\frac{b_1x^3}{x}+\frac{b_2y^3}{y}+\frac{b_3z^3}{z}}=

=\sqrt[3]{\frac{b_2y^3}{x}+\frac{b_2y^3}{y}+\frac{b_2y^3}{z}}=

=y \sqrt[3]{b_2} \sqrt[3]{\frac{1}{x}+\frac{1}{y}+\frac{1}{z}}= y \sqrt[3]{b_2} (*)

\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}= \dfrac{\sqrt[3]{b_1}}{y \sqrt[3]{b_2}}+\dfrac{\sqrt[3]{b_2}}{y \sqrt[3]{b_2}}+\dfrac{\sqrt[3]{b_3}}{y \sqrt[3]{b_2}}=1 \rightarrow{}

\rightarrow{} \sqrt[3]{b_1}+\sqrt[3]{b_2}+\sqrt[3]{b_3}=y \sqrt[3]{b_2} (**)

For (*) and (**):

\sqrt[3]{b_1x^2+b_2y^2+b_3z^2}=\sqrt[3] {b_1}+\sqrt[3] {b_2} + \sqrt[3] {b_3}

Regards.