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$a = 1 + {2^{1/3}} + {4^{1/3}}$, $\frac{1}{{{a^3}}} + \frac{3}{a} + \frac{3}{{{a^2}}} = ?$

 We have, $\frac{1}{{{a^3}}} + \frac{3}{a} + \frac{3}{{{a^2}}} + {1^3} = {\left( {\frac{1}{a} + 1} \right)^3}$

So, $\frac{1}{{{a^3}}} + \frac{3}{a} + \frac{3}{{{a^2}}} = {\left( {\frac{1}{a} + 1} \right)^3} - 1$

Now, $a = 1 + {2^{1/3}} + {4^{1/3}} = 1 + {2^{1/3}} + {({2^{1/3}})^2}$

Also, $({2^{1/3}} - 1)\left[ {{{({2^{1/3}})}^2} + {2^{1/3}} \times 1 + {1^2}} \right] = \left[ {{{({2^{1/3}})}^3} - {1^3}} \right] = 1$

So, $a = 1 + {2^{1/3}} + {4^{1/3}} = \frac{1}{{{2^{1/3}} - 1}}$

Thus, $\frac{1}{a} = {2^{1/3}} - 1$

Or, $\frac{1}{a} + 1 = {2^{1/3}}$

Or, ${\left( {\frac{1}{a} + 1} \right)^3} = 2$

Thus, $\frac{1}{{{a^3}}} + \frac{3}{a} + \frac{3}{{{a^2}}} = {\left( {\frac{1}{a} + 1} \right)^3} - 1 = 2 - 1 = 1$

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$f(x)=x^6+2x^4+x^3+2x+3 $

$\mathop {\lim }\limits_{x \to 1} \frac{{{x^n}f(1) - f(x)}}{{x - 1}} = 44$

$n=?$

Let $f(x)=x^6+2x^4+x^3+2x+3,x \in R $. Then the natural number n for which $\mathop {\lim }\limits_{x \to 1} \frac{{{x^n}f(1) - f(x)}}{{x - 1}} = 44$ is _ _ _ _ . Solution Since the limit has $\left[ {\frac{0}{0}} \right]$ form, L.H. Rule is applicable. Thus, $\mathop {\lim }\limits_{x \to 1} n{x^{n - 1}}f(1) - f'(x) = 44$ $\therefore nf(1) - f'(1) = 44$ $\therefore n.9 - ({6.1^5} + {8.1^3} + {3.1^2} + 2.1) = 44$ $ \Rightarrow 9n - 19 = 44$ $\Rightarrow n=7$