If n is the degree of the polynomial,left[ {f | vedclass

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(D) Let $P(x) = \left[\frac{1}{\sqrt{5 x^{3}+1}-\sqrt{5 x^{3}-1}}\right]^{8}+\left[\frac{1}{\sqrt{5 x^{3}+1}+\sqrt{5 x^{3}-1}}\right]^{8}$.Rationalizi

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$\left( {12,8{{\left( {10} \right)}^4}} \right)$

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(D) Let $P(x) = \left[\frac{1}{\sqrt{5 x^{3}+1}-\sqrt{5 x^{3}-1}}ight]^{8}+\left[\frac{1}{\sqrt{5 x^{3}+1}+\sqrt{5 x^{3}-1}}ight]^{8}$.Rationalizing the terms,we get:$P(x) = \left[\frac{\sqrt{5 x^{3}+1}+\sqrt{5 x^{3}-1}}{2}ight]^{8} + \left[\frac{\sqrt{5 x^{3}+1}-\sqrt{5 x^{3}-1}}{2}ight]^{8}$.$P(x) = \frac{1}{2^8} \left[ (\sqrt{5x^3+1} + \sqrt{5x^3-1})^8 + (\sqrt{5x^3+1} - \sqrt{5x^3-1})^8 ight]$.Using the expansion $(a+b)^8 + (a-b)^8 = 2 \sum_{k=0, 2, 4, 6, 8} \binom{8}{k} a^{8-k} b^k$,we have:$P(x) = \frac{2}{2^8} \left[ \binom{8}{0} (5x^3+1)^4 + \binom{8}{2} (5x^3+1)^3(5x^3-1) + \binom{8}{4} (5x^3+1)^2(5x^3-1)^2 + \binom{8}{6} (5x^3+1)(5x^3-1)^3 + \binom{8}{8} (5x^3-1)^4 ight]$.The highest power of $x$ is $x^{3 	imes 4} = x^{12}$,so $n = 12$.The coefficient of $x^{12}$ is given by $\frac{2}{2^8} 	imes 5^4 	imes \left[ \binom{8}{0} + \binom{8}{2} + \binom{8}{4} + \binom{8}{6} + \binom{8}{8} ight]$.Since $\sum_{k 	ext{ even}} \binom{8}{k} = 2^{8-1} = 2^7$,we have:$m = \frac{2}{2^8} 	imes 5^4 	imes 2^7 = \frac{2^8}{2^8} 	imes 5^4 	imes 2^3 = 8 	imes 10^4$.Thus,$(n, m) = (12, 8(10)^4)$.

If $n$ is the degree of the polynomial,$\left[ {\frac{1}{{\sqrt {5{x^3} + 1} - \sqrt {5{x^3} - 1} }}} \right]^8 + \left[ {\frac{1}{{\sqrt {5{x^3} + 1} + \sqrt {5{x^3} - 1} }}} \right]^8$ and $m$ is the coefficient of $x^{12}$ in it,then the ordered pair $(n, m)$ is equal to

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