prove the identity

Albert

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

Last edited:

mente oscura

Well-known member
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.

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

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

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

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

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

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

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.