Question

please answer this question​

Answer 1

Topic

Exponent Laws

• Roots

A number, $a$ to the n-th power is $a^{n}$.

Let the power be $k$, then

$\implies k=a^{n}$

$\implies \sqrt[n]{k} =a$

However,

$\implies (\sqrt[n]{k} )^{n}=a^{n}$

So,

$\implies (\sqrt[n]{k} )^{n}=k$

For this to be true,

$\implies (k^{\frac{1}{n} })^{n}=k$

Therefore, a new mathematical convention is

$\implies \sqrt[n]{k} =k^{\frac{1}{n} }$

And this is the exponent rule for roots.

Solution

Given Number

$=\sqrt[4]{256} +\sqrt{8} -3\sqrt{2} -\sqrt[3]{343}$

$=\sqrt[4]{2^{8} }+\sqrt{2^{3}} -3\sqrt{2} -\sqrt[3]{7^{3}}$

$=2^\frac{8}{4} +2^{\frac{3}{2} }-3\sqrt{2} -(7^{3})^{\frac{1}{3} }$

$=2^{2}+2\sqrt{2} -3\sqrt{2} -7$

$=-3-\sqrt{2}$

This is the required answer.