Question

prove it...
solve it correctly....explanation must...​

Answer 1

Answer:

Heya buddy....

Here is ur answer...

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Answer 2

To Prove:

$\mathrm{\sqrt[\big{(p+q)}]{\dfrac{x^{p^{2}}}{x^{q^{2}}}}\times \sqrt[\big{(q+r)}]{\dfrac{x^{q^{2}}}{x^{r^{2}}}}\times \sqrt[\big{(r+p)}]{\dfrac{x^{r^{2}}}{x^{p^{2}}}}=1}$

Solution:

We know that,

• $\mathrm{\sqrt[\big{n}]{x}=x^{\frac{1}{n}}}$

• $(a^{m})^{n}=a^{mn}$

• $\mathrm{\dfrac{a^{m}}{a^{n}}=a^{m-n}}$

• $\mathrm{a^{2}-b^{2}=(a+b)(a-b)}$

• $\mathrm{a^{m}\times a^{n}=a^{m+n}}$

Now,

$\sf{L.H.S.=\sqrt[\big{(p+q)}]{\dfrac{x^{p^{2}}}{x^{q^{2}}}}\times \sqrt[\big{(q+r)}]{\dfrac{x^{q^{2}}}{x^{r^{2}}}}\times \sqrt[\big{(r+p)}]{\dfrac{x^{r^{2}}}{x^{p^{2}}}}}$

$\rightarrow\sf{\bigg(\dfrac{x^{p^{2}}}{x^{q^{2}}}\bigg)^{\frac{1}{p+q}}\times \bigg(\dfrac{x^{q^{2}}}{x^{r^{2}}}\bigg)^{\frac{1}{q+r}}\times \bigg(\dfrac{x^{r^{2}}}{x^{p^{2}}}\bigg)^{\frac{1}{r+p}}}$

$\rightarrow\sf{\Big(x^{p^{2}-q^{2}}\Big)^{\frac{1}{p+q}}\times \Big(x^{q^{2}-r^{2}}\Big)^{\frac{1}{q+r}}\times \Big(x^{r^{2}-p^{2}}\Big)^{\frac{1}{r+p}}}$

$\rightarrow\sf{\bigg(x^{\frac{p^{2}-q^{2}}{p+q}}\bigg)\times \bigg(x^{\frac{q^{2}-r^{2}}{q+r}}\bigg)\times \bigg(x^{\frac{r^{2}-p^{2}}{r+p}}\bigg)}$

$\rightarrow\sf{\bigg(x^{\frac{\cancel{(p+q)}(p-q)}{\cancel{p+q}}}\bigg)\times \bigg(x^{\frac{\cancel{(q+r)}(q-r)}{\cancel{q+r}}}\bigg)\times \bigg(x^{\frac{\cancel{(r+p)}(r-p)}{\cancel{r+p}}}\bigg)}$

$\rightarrow\sf{\Big(x^{p-q}\Big)\times \Big(x^{q-r}\Big)\times \Big(x^{r-p}\Big)}$

$\rightarrow\sf{x^{p-q+q-r+r-p}}$

$\rightarrow\sf{x^{0}}$

$\rightarrow\sf{1}$

$\rightarrow\sf{R.H.S.}$

Since, L.H.S.=R.H.S.

Hence Proved