Question

Show that cube root of ax^2+by^2+cz^2 is equal to cube root of a + cube rootof b+ cube root of c if ax^3=by^3=cz^3 and xy+yz+zx=xzy

Answer 1

We have to prove:

$\sqrt[3]{ax^{2}+by^{2}+cz^{2}}=\sqrt[3]{a}+\sqrt[3]{b}+\sqrt[3]{c}$

Solution:

Given,

$\quad ax^{3}=by^{3}=cz^{3}$

$\to ax^{3}=by^{3}=cz^{3}=k\:(\neq 0)$ (say)

Then,

$\quad\quad x=\big(\frac{k}{a}\big)^{1/3}$

$\quad\quad y=\big(\frac{k}{b}\big)^{1/3}$

$\quad\quad z=\big(\frac{k}{c}\big)^{1/3}$

Substituting the above values in the given relation

$\quad xy+yz+zx=xzy$

$\Rightarrow \frac{1}{z}+\frac{1}{x}+\frac{1}{y}=1$, we get

$\quad \frac{1}{\big(\frac{k}{c}\big)^{1/3}}+\frac{1}{\big(\frac{k}{b}\big)^{1/3}}+\frac{1}{\big(\frac{k}{a}\big)^{1/3}}=1$

$\Rightarrow \big(\frac{c}{k}\big)^{1/3}+\big(\frac{b}{k}\big)^{1/3}+\big(\frac{c}{k}\big)^{1/3}=1$

$\Rightarrow k^{1/3}=a^{1/3}+b^{1/3}+c^{1/3}$

So, we have

$\quad x=\frac{a^{1/3}+b^{1/3}+c^{1/3}}{a^{1/3}}$

$\Rightarrow x^{2}=\big(\frac{a^{1/3}+b^{1/3}+c^{1/3}}{a^{1/3}}\big)^{2}$

$\Rightarrow ax^{2}=a\big(\frac{a^{1/3}+b^{1/3}+c^{1/3}}{a^{1/3}}\big)^{2}$

$\Rightarrow ax^{2}=a^{1/3}(a^{1/3}+b^{1/3}+c^{1/3})^{2}$

$\quad y=\frac{a^{1/3}+b^{1/3}+c^{1/3}}{b^{1/3}}$

$\Rightarrow y^{2}=\big(\frac{a^{1/3}+b^{1/3}+c^{1/3}}{b^{1/3}}\big)^{2}$

$\Rightarrow by^{2}=b\big(\frac{a^{1/3}+b^{1/3}+c^{1/3}}{b^{1/3}}\big)^{2}$

$\Rightarrow by^{2}=b^{1/3}(a^{1/3}+b^{1/3}+c^{1/3})^{2}$

$\quad z=\frac{a^{1/3}+b^{1/3}+c^{1/3}}{c^{1/3}}$

$\Rightarrow z^{2}=\big(\frac{a^{1/3}+b^{1/3}+c^{1/3}}{c^{1/3}}\big)^{2}$

$\Rightarrow cz^{2}=c\big(\frac{a^{1/3}+b^{1/3}+c^{1/3}}{c^{1/3}}\big)^{2}$

$\Rightarrow cz^{2}=c^{1/3}(a^{1/3}+b^{1/3}+c^{1/3})^{2}$

Now, $ax^{2}+by^{2}+cz^{2}$

$=(a^{1/3}+b^{1/3}+c^{1/3})(a^{1/3}+b^{1/3}+c^{1/3})^{2}$

$=(a^{1/3}+b^{1/3}+c^{1/3})^{3}$

$\Rightarrow ax^{2}+by^{2}+cz^{2}=(a^{1/3}+b^{1/3}+c^{1/3})^{3}$

Taking cube root to both sides, we get

$\quad (ax^{2}+by^{2}+cz^{2})^{1/3}=[(a^{1/3}+b^{1/3}+c^{1/3})^{3}]^{1/3}$

$\Rightarrow (ax^{2}+by^{2}+cz^{2})^{1/3}=a^{1/3}+b^{1/3}+c^{1/3}$

$\Rightarrow \sqrt[3]{ax^{2}+by^{2}+cz^{2}}=\sqrt[3]{a}+\sqrt[3]{b}+\sqrt[3]{c}$

Hence, proved.

Answer 2

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