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

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The discussion focuses on proving the identity $\sqrt[3]{b_1x^2+b_2y^2+b_3z^2}=\sqrt[3]{b_1}+\sqrt[3]{b_2}+\sqrt[3]{b_3}$ under the conditions $b_1x^3=b_2y^3=b_3z^3$ and $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=1$. Participants emphasize the importance of these equations in establishing the relationship between the variables and constants involved. The proof hinges on manipulating the given equations to derive the desired equality definitively.

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

Albert said:
(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.

\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.
 

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