Question

Solve:-
$\frac{(7 {p}^{2} {q}^{9} {r}^{5} {)}^{2}(4pqr) {}^{3} }{(14 {p}^{6} {q}^{10} {r}^{4} {)}^{2} }$
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Answer 1

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Answer 2

$\large\underline{\sf{Solution-}}$

$\rm :\longmapsto\:\dfrac{(7 {p}^{2} {q}^{9} {r}^{5} {)}^{2}(4pqr) {}^{3} }{(14 {p}^{6} {q}^{10} {r}^{4} {)}^{2} }$

Consider,

$\rm :\longmapsto\:(7 {p}^{2} {q}^{9} {r}^{5} {)}^{2}$

We know,

$\boxed{ \bf{ {(xy)}^{n} \: = \: {x}^{n} {y}^{n}}}$

So, using this

$\rm :\longmapsto\: {7}^{2} {p}^{4} {q}^{18} {r}^{10}$

Hence,

$\rm :\longmapsto\:(7 {p}^{2} {q}^{9} {r}^{5} {)}^{2} = {7}^{2} {p}^{4} {q}^{18} {r}^{10}$

Consider,

$\rm :\longmapsto\:(4pqr) {}^{3} = {4}^{3} {p}^{3} {q}^{3} {r}^{3}$

Consider,

$\rm :\longmapsto\:{(14 {p}^{6} {q}^{10} {r}^{4} {)}^{2} }$

$\rm \: \: = \: \:{(7 \times 2 \times {p}^{6} {q}^{10} {r}^{4} {)}^{2} }$

$\rm \: = \: \: {7}^{2} {2}^{2} {p}^{12} {q}^{20} {r}^{8}$

Hence,

$\rm :\longmapsto\:\dfrac{(7 {p}^{2} {q}^{9} {r}^{5} {)}^{2}(4pqr) {}^{3} }{(14 {p}^{6} {q}^{10} {r}^{4} {)}^{2} }$

$\rm \: \: = \:\dfrac{ {7}^{2} {p}^{4} {q}^{18} {r}^{10} \times {4}^{3} {p}^{3} {q}^{3} {r}^{3} }{{7}^{2} {2}^{2} {p}^{12} {q}^{20} {r}^{8}}$

We know,

$\boxed{ \bf{ {x}^{m} \times {x}^{n} \: = \: {x}^{m + n} }}$

So using this, in numerator, we get

$\rm \: \: = \:\dfrac{ {7}^{2} {p}^{4 + 3} {q}^{18 + 3} {r}^{10 + 3} \times {(2 \times 2)}^{3} }{{7}^{2} {2}^{2} {p}^{12} {q}^{20} {r}^{8}}$

$\rm \: \: = \:\dfrac{ \cancel {7}^{2} {p}^{7} {q}^{21} {r}^{13} \times {(2)}^{6} }{ \cancel{7}^{2} {2}^{2} {p}^{12} {q}^{20} {r}^{8}}$

We know,

$\boxed{ \bf{ {x}^{m} \div {x}^{n} \: = \: {x}^{m - n} }}$

$\rm \: \: = \: {p}^{7 - 12} {q}^{21 - 20} {r}^{13 - 8} {(2)}^{6 - 2}$

$\rm \: \: = \: {p}^{ - 5} {q}^{1} {r}^{5} {(2)}^{4}$

$\rm \: \: = \:\dfrac{16q {r}^{5} }{ {p}^{5} }$

Hence,

$\bf :\longmapsto\:\dfrac{(7 {p}^{2} {q}^{9} {r}^{5} {)}^{2}(4pqr) {}^{3} }{(14 {p}^{6} {q}^{10} {r}^{4} {)}^{2} } \: \: = \:\dfrac{16q {r}^{5} }{ {p}^{5} }$

Additional Information :-

$\boxed{ \bf{ {x}^{0} \: = \: 1}}$

$\boxed{ \bf{ {x}^{ - y} \: = \: \frac{1}{ {x}^{y} } }}$

$\boxed{ \bf{ {x}^{m} \times {y}^{m} = {(xy)}^{m} }}$