Question

pls answer the question immediately.... . I will give brainlist answer​

Answer 1

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

a)

Given:

$\sf = \bigg(\dfrac{4}{3} \bigg)^{ - 3}$

$\sf = \dfrac{ {4}^{ - 3} }{ {3}^{ - 3} }$

As we know that:

$\sf \implies {x}^{ - a} = \dfrac{1}{ {x}^{a} }$

We can write the expression as:

$\sf = \dfrac{ {3}^{3} }{ {4}^{3} }$

$\sf = \bigg(\dfrac{3}{4} \bigg)^{3} \: \: \: (Answer)$

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b)

Given:

$\sf = \bigg \{ { \bigg( \dfrac{1}{5} \bigg)}^{ - 2} \bigg\}^{3}$

As we know that:

$\sf \implies ({x}^{a})^{b} = {x}^{ab}$

We get:

$\sf = \bigg( { \dfrac{1}{5} \bigg)}^{ - 2 \times 3}$

$\sf = \bigg( { \dfrac{1}{5} \bigg)}^{ -6}$

$\sf = {5}^{6} \: \: \: (Answer)$

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c)

Given:

$\sf = \bigg \{ { \bigg( \dfrac{3}{7} \bigg)}^{ - 3} \bigg \}^{ - 4}$

As we know that:

$\sf \implies ({x}^{a})^{b} = {x}^{ab}$

We get:

$\sf = { \bigg( \dfrac{3}{7} \bigg)}^{( - 3) \times ( - 4)}$

$\sf = { \bigg( \dfrac{3}{7} \bigg)}^{12} \: \: \: (Answer)$

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d)

Given:

$\sf = {8}^{ - 5} \times {8}^{3} \times {8}^{ - 4}$

As we know that:

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

We get:

$\sf = {8}^{ - 5 + 3 - 4}$

$\sf = {8}^{ - 6}$

$\sf = \dfrac{1}{ {8}^{6} } \: \: \: (Answer)$

$\texttt{\textsf{\large{\underline{Know More}:}}}$

If a, b are positive real numbers and m, n are rational numbers, then the following results hold -

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

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

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

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

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

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

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

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

Answer 2

$\pink{ \boxed{\boxed{\begin{array}{cc}\bf \underline{ \maltese\:\:solution-\:a:\:\maltese}\\\\ \bf \: {( \frac{4}{3} })^{ - 3} \\ \\ = \{{ {( \frac{4}{3} })^{3} } \}^{ - 1} \\ \\ = \frac{1}{ {( \frac{4}{3} })^{3} } \: \: \: \red{ \{ \because \: \bf \: {x}^{ - 1} = \frac{1}{x} \} } \\ \\ = ( \frac{3}{4} ) ^{3} \: \: \: \red{ \{ \because \bf \: \frac{1}{ \frac{a}{b} } = \frac{b}{a} \} }\: \end{array}}}}$

$\orange{ \boxed{\boxed{\begin{array}{cc}\bf \underline{ \maltese\:\:solution-\:b:\:\maltese}\\\\ \bf \: \{ {( \frac{1}{5} })^{ - 2} \} ^{3} \\ \\ = {( \frac{1}{5} })^{ - 2 \times 3} \: \: \: \red{ \{ \because \bf \: \{{(x) ^{m} } \}^{n} = {x}^{mn} \} } \\ \\ = {( \frac{1}{5} })^{ - 6} \\ \\ = {5}^{6} \: \: \end{array}}}}$

${ \boxed{\boxed{\begin{array}{cc} \bf \underline{ \maltese\:\:solution-\:c:\:\maltese}\\\\ \bf \:\{ {( \frac{3}{7} })^{ - 3} \} ^{ - 4} \\ \\ = {( \frac{3}{7} })^{ - 3 \times( - 4)} \: \: \: \red{ \{ \because \bf \: \{{(x) ^{m} } \}^{n} = {x}^{mn} \} } \\ \\ = {( \frac{3}{7} })^{ 12} \\ \\ \: \end{array}}}}$

$\small{\blue{ \boxed{\boxed{\begin{array}{cc} \bf \underline{ \maltese\:\:soution-\:d:\:\maltese}\\\\ \bf \: {(8)}^{ - 5} \times ( {8})^{3} \times {(8)}^{ - 4} \\ \\ = {8}^{ - 5 + 3 + ( - 4)} \: \: \: \red{ \{ \because \bf \: {x}^{m} \times {x}^{n} \times ... \times {x}^{r} = {x}^{m + n + ... + r} \} }\: \\ \\ = {8}^{ - 5 + 3 - 4} \\ \\ = {8}^{ - 6} \\ \\ = \frac{1}{ {8}^{6} } \: \end{array}}}} }$