Question

Evaluate

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Answer 1

$\rm\large\underline{\text{Basics}}$

$\cdots\longrightarrow\textrm{ (a+b)(a-b)=a^2-b^2 as a polynomial identity.}$

$\cdots\longrightarrow\textrm{ a^{\frac{m}{n}}=\sqrt[n]{a^{m}} as a power rule.}$

$\cdots\longrightarrow\textrm{ (ab)^{r}=a^rb^r as a power rule.}$

$\cdots\longrightarrow\textrm{ (a^r)^{s}=a^{rs} as a power rule.}$

$\rm\large\underline{\text{Solution}}$

$\begin{aligned}\cdots\longrightarrow (14^2-13^2)^{\frac{2}{3}}&=(14+13)^{\frac{2}{3}}\times (14-13)^{\frac{2}{3}}\\&=27^{\frac{2}{3}}\times1^{\frac{2}{3}}\\&=(3^3)^{\frac{2}{3}}\\&=3^2\\&=9\end{aligned}$

$\begin{aligned}\cdots\longrightarrow (15^2-12^2)^{\frac{1}{4}}&=(15+12)^{\frac{1}{4}}\times (15-12)^{\frac{1}{4}}\\&=27^{\frac{1}{4}}\times3^{\frac{1}{4}}\\&=(3^4)^{\frac{1}{4}}\\&=3^1\\&=3\end{aligned}$

The required value is,

$\cdots\longrightarrow\dfrac{(14^2-13^2)^{\frac{2}{3}}}{(15^2-12^2)^{\frac{1}{4}}}=\dfrac{9}{3}=3$

$\rm\large\underline{\text{Question review!}}$

(Question)

Find the value of $\sqrt{\dfrac{2^{-10}+2^{-8}+2^{-6}}{2^{10}+2^8+2^6}}$.

(Solution)

Let us first factorize by the identity $ka+kb=k(a+b)$.

$\begin{aligned}\cdots\longrightarrow\sqrt{\dfrac{2^{-10}\times (2^{0}+2^{2}+2^{4})}{2^6\times (2^{4}+2^{2}+2^{0})}}&=\sqrt{\dfrac{2^{-10}}{2^{6}}}\\&=\sqrt{2^{-16}}\\&=2^{-8}\\&=\dfrac{1}{256}\end{aligned}$

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Answer 2

$\large\underline{ \underline{ \rm{Explanation:}}}$

To Find,

• $\text{The value of} \: \frac{ {( {14}^{2} - {13}^{2}) }^{ \frac{2}{3} } }{ {( {15}^{2} - {12}^{2} ) }^{ \frac{1}{4} } }.$

Solving the problem,

$\cdot \cdot \cdot \longrightarrow \frac{ {( {14}^{2} - {13}^{2}) }^{ \frac{2}{3} } }{ {( {15}^{2} - {12}^{2} ) }^{ \frac{1}{4} } } = \frac{ {( {14}^{2} - {13}^{2}) }^{ \frac{2}{3} } }{ {( {15}^{2} - {12}^{2} ) }^{ \frac{1}{4} } }$

As we know that,

$[ {x}^{2} - {y}^{2} = (x + y)(x - y) ]$

We get,

$\cdot \cdot \cdot \longrightarrow \frac{ {( {14}^{2} - {13}^{2}) }^{ \frac{2}{3} } }{ {( {15}^{2} - {12}^{2} ) }^{ \frac{1}{4} } } = \frac{ { \{(14 + 13)(14 - 13) \}}^{ \frac{2}{3} } }{ { \{(15 + 12)(15 - 12) \}}^{ \frac{1}{4} } }$

$\cdot \cdot \cdot \longrightarrow \frac{ {( {14}^{2} - {13}^{2}) }^{ \frac{2}{3} } }{ {( {15}^{2} - {12}^{2} ) }^{ \frac{1}{4} } } = \frac{ { \{1(27) \}}^{ \frac{2}{3} } }{ { \{3(27)\}}^{ \frac{1}{4} } }$

$\cdot \cdot \cdot \longrightarrow \frac{ {( {14}^{2} - {13}^{2}) }^{ \frac{2}{3} } }{ {( {15}^{2} - {12}^{2} ) }^{ \frac{1}{4} } } = \frac{ { (27)}^{ \frac{2}{3} } }{ {(81)}^{ \frac{1}{4} } }$

$\cdot \cdot \cdot \longrightarrow \frac{ {( {14}^{2} - {13}^{2}) }^{ \frac{2}{3} } }{ {( {15}^{2} - {12}^{2} ) }^{ \frac{1}{4} } } = \frac{ {( {27}^{ \frac{1}{3} } )}^{2} }{ \sqrt[4]{81} }$

$\cdot \cdot \cdot \longrightarrow \frac{ {( {14}^{2} - {13}^{2}) }^{ \frac{2}{3} } }{ {( {15}^{2} - {12}^{2} ) }^{ \frac{1}{4} } } = \frac{ ({ \sqrt[3]{27} )}^{2} }{ \sqrt[4]{3 \times 3 \times 3 \times 3} }$

$\cdot \cdot \cdot \longrightarrow \frac{ {( {14}^{2} - {13}^{2}) }^{ \frac{2}{3} } }{ {( {15}^{2} - {12}^{2} ) }^{ \frac{1}{4} } } = \frac{ ({ \sqrt[3]{3 \times 3 \times 3} )}^{2} }{ 3}$

$\cdot \cdot \cdot \longrightarrow \frac{ {( {14}^{2} - {13}^{2}) }^{ \frac{2}{3} } }{ {( {15}^{2} - {12}^{2} ) }^{ \frac{1}{4} } } = \frac{ ({3 )}^{2} }{ 3}$

$\cdot \cdot \cdot \longrightarrow \frac{ {( {14}^{2} - {13}^{2}) }^{ \frac{2}{3} } }{ {( {15}^{2} - {12}^{2} ) }^{ \frac{1}{4} } } = \frac{ \cancel{3 } \times 3 }{ \cancel{3}}$

$\cdot \cdot \cdot \longrightarrow \boxed{\frac{ {( {14}^{2} - {13}^{2}) }^{ \frac{2}{3} } }{ {( {15}^{2} - {12}^{2} ) }^{ \frac{1}{4} } } = 3}$

$\large\underline{ \underline{ \rm{Required \: Answer:}}}$

• $\boxed{\frac{ {( {14}^{2} - {13}^{2}) }^{ \frac{2}{3} } }{ {( {15}^{2} - {12}^{2} ) }^{ \frac{1}{4} } } = 3}$

$\large\underline{ \underline{ \text{More Information:}}}$

• $\sqrt[n]{a} = {a}^{ \frac{1}{n} }$

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

• $\frac{ {a}^{m} }{ {a}^{n} } = {a}^{m - n}$

• ${a}^{m} \times {b}^{m} = {(ab)}^{m}$