Show that: [tex]\sqrt[ab]{ \frac{ {x}^{a} }{ {x}^{b} } } . \sqrt[bc]{ \frac{ {x}^{b} }{ {x}^{c} } } . \sqrt[ca]{ \frac{ {x}^{c} }{ {x}^{a} } } = 1.[/tex]

Brainliest! Brainliest..ans.
​

Here we need to Prove that ,

[tex]\bf\implies \sqrt[ab]{ \frac{ {x}^{a} }{ {x}^{b} } } . \sqrt[bc]{ \frac{ {x}^{b} }{ {x}^{c} } } . \sqrt[ca]{ \frac{ {x}^{c} }{ {x}^{a} } } = 1.[/tex]

Lets start with LHS , which is ,

[tex]\bf\implies LHS = \sqrt[ab]{ \dfrac{ {x}^{a} }{ {x}^{b} } } . \sqrt[bc]{ \dfrac{ {x}^{b} }{ {x}^{c} } } . \sqrt[ca]{ \dfrac{ {x}^{c} }{ {x}^{a} } } [/tex]

We know that ,

[tex]\bf\implies \sqrt[n]{k^m }= k^{\frac{m}{n}}[/tex]

• So that ,

[tex]\bf\implies \sqrt[ab]{x^{a-b}} . \sqrt[bc]{x^{b-c}} .\sqrt[ac]{x^{c-a}} \\\\\bf\implies x^{\frac{a-b}{ab}} . x ^{\frac{b-c}{bc}} .x^{\frac{c-a}{ca}} \\\\\bf\implies x^{\frac{ ac - bc + ab - ac + bc - ba }{abc } }\\\\\bf\implies x^{\frac{0}{abc}} \\\\\bf\implies x^0 \\\\\bf\implies 1 = RHS [/tex]

Hence ,

[tex]\implies \boxed{\bf \red{ LHS = RHS }} [/tex]

[tex]\large\underline{\sf{Solution-}}[/tex]

Consider LHS:

[tex]\sf= \sqrt[\sf ab]{\dfrac{\sf x^{a}}{\sf x^{b}}}\cdot \sqrt[\sf bc]{\dfrac{\sf x^{b}}{\sf x^{c}}}\cdot \sqrt[\sf ac]{\dfrac{\sf x^{c}}{\sf x^{a}}}[/tex]

As we know that:

[tex]\sf\red\leadsto\dfrac{x^{a}}{x^{b}}=x^{a-b}[/tex]

We get:

[tex]\sf= \sqrt[\sf ab]{\sf x^{a-b}}\cdot \sqrt[\sf bc]{\sf x^{b-c}}\cdot \sqrt[\sf ac]{\sf x^{c-a}}[/tex]

We can also write it as:

[tex]\sf= x^{\dfrac{a-b}{ab}}\cdot x^{\dfrac{b-c}{bc}}\cdot x^{\dfrac{c-a}{ac}}[/tex]

We know that:

[tex]\sf \red\leadsto x^{a}\cdot x^{b}=x^{a+b}[/tex]

We get:

[tex]\sf= x^{\bigg(\dfrac{a-b}{ab}+\dfrac{b-c}{bc}+\dfrac{c-a}{ac}\bigg)}[/tex]

[tex]\sf= x^{\bigg(\dfrac{c(a-b)+a(b-c)+b(c-a)}{abc}\bigg)}[/tex]

[tex]\sf= x^{\bigg(\dfrac{ac-bc+ab-ac+bc-ab}{abc}\bigg)}[/tex]

[tex]\sf= x^{\bigg(\dfrac{0}{abc}\bigg)}[/tex]

[tex]\sf= x^{0}[/tex]

[tex]\sf=1[/tex]

[tex]\textsf{= RHS}\\[/tex]

<1 and <2 are complementary. Solve for the measure of <1.

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