- #1
Albert1
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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:
$\sqrt[3]{b_1x^2+b_2y^2+b_3z^2}=\sqrt[3] {b_1}+\sqrt[3] {b_2} + \sqrt[3] {b_3}$
$b_1x^3=b_2y^3=b_3z^3$
(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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