Question

$\sf{if \quad {a}^{p} = {b}^{q} = {c}^{r} = abc } \\ \bf{then} \: \: \: \green{\boxed{ \bf{ \orange{pqr}}} }=$
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Answer 1

$\huge{\red{\bf{answer}}}$

we have a^p = a^q = c^r = abc, so

${a}^{p} = abc \\ \\ a = (abc)\frac{1}{p} \: \: \: \: \: \: \: .1 \\ \\ \\ {a}^{q} = abc \\ \\ b = (abc)\frac{1}{q} \: \: \: \: \: .2 \\ \\ and \\ \\ {c}^{r} = abc \\ \\ {c}^{r} =(abc) \frac{1}{r}\: \: \: \: \: .3$

Now we multiple equation 1 , 2 and 3 and get;

$abc = (abc) \frac{1}{p} + \frac{1}{q} + \frac{1}{c}$

( As we know (a^m) (a^n) = (a^m+n))

Be comparing both we side get;

$\frac{1}{p} + \frac{1}{q} + \frac{1}{r} = 1$

$= \frac{qr + pr + pq}{pqr} = 1$

answer : pqr = qr + pr + pq

Answer 2

$\underline{\underline{\bf{Given-}}}$

$\sf{if \quad {a}^{p} = {b}^{q} = {c}^{r} = abc }$

$\underline{\underline{\bf{To\:Find-}}}$

$\sf{Value \: of \: PQR }$

$\underline{\underline{\bf{Solution-}}}$

$\sf{a^p = abc}$

$\sf a = (abc)^ {\frac{1}{p}} \_ \: \_ \: \_ \: \_ \: \_ (1)$

$\sf \quad \quad \quad \quad \quad \quad \: (since \: after \: transfering \: we \: will \: take \: the \: pth \: root )$

$\tt{Similarly-}$

$\sf{b^q = abc}$

$\sf{b = (abc)^{\frac{1}{q}}\_ \: \_ \: \_ \: \_ \: \_ \: (2)}$

$\tt{and}$

$\sf{c^r = abc}$

$\sf{c = (abc)^{\frac{1}{r}}\_ \: \_ \: \_ \: \_ \: \_ \: (3)}$

$\underline{ \sf{Multiplying \: eq. (1),(2) and (3)}}$

$\sf{abc = (abc)^{\frac{1}{p}}.(abc)^{\frac{1}{q}}.(abc)^{\frac{1}{r}}}$

$\sf{abc = (abc)^{\frac{1}{p} +\frac{1}{q}+\frac{1}{r}}}$

$\sf \quad \quad \quad \quad \quad \quad \: (a^m \times a^n = a^{m+n} )$

$\underline{ \sf{Comparing \: powers \: of \: both \: sides-}}$

$\sf1 = \dfrac{1}{p}+\dfrac{1}{q}+\dfrac{1}{r}$

$\sf{1 = \dfrac{qr+pr+pq}{pqr} }$

$\boxed{\bf{pqr = qr+pr+pq}}$