Question

The sum is shown in the picture
(PROVE LHS = RHS)
​

Answer 1

$\textsf{\large{\underline{Solution}:}}$

We have to prove:

$\sf\implies \sqrt[\sf ab]{\dfrac{\sf x^{a}}{\sf x^{b}}}\cdot \sqrt[\sf bc]{\dfrac{\sf x^{b}}{\sf x^{c}}}\cdot \sqrt[\sf ac]{\dfrac{\sf x^{c}}{\sf x^{a}}}=1$

Taking LHS, we get:

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

As we know that:

$\sf\implies \dfrac{x^{a}}{x^{b}}=x^{a-b}$

We get:

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

We can also write it as:

$\sf= x^{\dfrac{a-b}{ab}}\cdot x^{\dfrac{b-c}{bc}}\cdot x^{\dfrac{c-a}{ac}}$

We know that:

$\sf\implies x^{a}\cdot x^{b}=x^{a+b}$

We get:

$\sf= x^{\bigg(\dfrac{a-b}{ab}+\dfrac{b-c}{bc}+\dfrac{c-a}{ac}\bigg)}$

$\sf= x^{\bigg(\dfrac{c(a-b)+a(b-c)+b(c-a)}{abc}\bigg)}$

$\sf= x^{\bigg(\dfrac{ac-bc+ab-ac+bc-ab}{abc}\bigg)}$

$\sf= x^{\bigg(\dfrac{0}{abc}\bigg)}$

$\sf= x^{0}$

$\sf=1$

We observe that LHS = RHS..!!

Hence Proved.

$\texttt{\textsf{\large{\underline{Additionals}:}}}$

$\sf 1. \: \: {a}^{m} \times {a}^{n} = {a}^{m + n}$

$\sf 2. \: \: ({a}^{m})^{n} = {a}^{mn}$

$\sf 3. \: \: \dfrac{ {a}^{m} }{ {a}^{n} } = {a}^{m - n}$

$\sf4. \: \: {a}^{m} \times {b}^{m} = {(ab)}^{m}$

$\sf5. \: \: \bigg(\dfrac{a}{b} \bigg)^{m} = \dfrac{ {a}^{m} }{ {b}^{m} }$

$\sf6. \: \: {a}^{ - n} = \dfrac{1}{ {a}^{n} }$

$\sf7. \: \: {a}^{n} = {b}^{n} \rightarrow a = b, n \neq0$

$\sf8. \: \: {a}^{m} = {a}^{n} \rightarrow m = n, a \neq 1$

Answer 2

We have to prove that,

$\large \bf \sqrt[ab]{ \frac{ {x}^{a} }{ {x}^{b} } } . \sqrt[bc]{ \frac{ {x}^{b} }{ {x}^{c} } } . \sqrt[ca]{ \frac{ {x}^{c} }{ {x}^{a} } } = 1$

LHS -

Using formula,

$\frac{ {x}^{m} }{ {x}^{n} } = {x}^{m - n}$

We get,

$\bf \sqrt[ab]{ {x}^{a - b} } . \sqrt[bc]{ {x}^{bc} } . \sqrt[ca]{ {x}^{c - a} }$

Now, we use the formula,

$\sqrt[m]{ {x}^{n} } = {x}^{ \frac{n}{m} }$

Then,

$\bf {x}^{ \frac{a - b}{ab} } . { x }^{ \frac{b - c}{bc} } . {x}^{ \frac{c - a}{ca} }$

Using formula,

${x}^{m} \times {x}^{n} = {x}^{m + n}$

We can write,

$\bf{x}^{ \frac{a - b}{ab} + \frac{b - c}{bc} + \frac{c - a}{ca} }$

$= > \bf {x}^{ \small( \frac{c(a - b) + a(b - c) + b(c - a)}{abc} \small) } \\ \\ = > \bf {x}^{ (\frac{ca - bc + ab - ac + bc - ab}{abc} )} \\ \\ = > \bf {x}^{ \frac{0}{abc} } \\ \\ \bf = > {x}^{0} = 1$

=> LHS = RHS

Hence proved.

Hope it helps.