यदि n, बहुपद {left[ {frac{1}{{sqrt {5{x^3} + | vedclass

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Question

$\left[\frac{1}{\sqrt{5 x^{3}+1}-\sqrt{5 x^{3}-1}}\right]^{8}+\left[\frac{1}{\sqrt{5 x^{3}+1}+\sqrt{5 x^{3}-1}}\right]^{8}$

After rationalise the p

Vedclass · Accepted Answer

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

Vedclass · Answer

$\left[\frac{1}{\sqrt{5 x^{3}+1}-\sqrt{5 x^{3}-1}}ight]^{8}+\left[\frac{1}{\sqrt{5 x^{3}+1}+\sqrt{5 x^{3}-1}}ight]^{8}$

After rationalise the polynomial we get

=$\left[\frac{1}{\sqrt{5 x^{3}+1}-\sqrt{5 x^{3}-1}} 	imes \frac{\sqrt{5 x^{3}+1}+\sqrt{5 x^{3}-1}}{\sqrt{5 x^{3}+1}+\sqrt{5 x^{3}-1}}ight]^{8}$

$+\left[\frac{1}{\sqrt{5 x^{3}+1}+\sqrt{5 x^{3}-1}} 	imes \frac{\sqrt{5 x^{3}+1}-\sqrt{5 x^{3}-1}}{\sqrt{5 x^{3}+1}-\sqrt{5 x^{3}-1}}ight]^{8}$

=$\left[\frac{\sqrt{5 x^{3}+1}+\sqrt{5 x^{3}-1}}{\left(5 x^{3}+1ight)-\left(5 x^{3}-1ight)}ight]^{8}+\left[\frac{\sqrt{5 x^{3}+1}-\sqrt{5 x^{3}-1}}{\left(5 x^{3}+1ight)-\left(5 x^{3}-1ight)}ight]^{8}$

$=\frac{1}{2^{8}}\left[(\sqrt{5 x^{3}+1}+\sqrt{5 x^{3}-1})^{8}+(\sqrt{5 x^{3}+1}-\sqrt{5 x^{3}-1})^{8}ight]$

=$^{8} C_{0}(\sqrt{5 x^{3}+1})^{8}+^{8} C_{2}(\sqrt{5 x^{3}+1})^{6}(\sqrt{5 x^{3}-1})^{2}$$\frac{1}{2^{8}} \cdot C_{4}(\sqrt{5 x^{3}+1})^{4}(\sqrt{5 x^{3}-1})^{4}+$

$^{\prime} C_{6}\left(\sqrt{5 x^{3}+1}^{2}(\sqrt{5 x^{3}})^{6}+^{6} C_{8}\left(y^{\sqrt{5 x^{3}-1}}ight.ight.$

$\frac{1}{2^{8}}\left[\begin{array}{l}{^{8} C_{0}\left(5 x^{3}+1ight)^{4}+^{8} C_{2}\left(5 x^{3}+1ight)^{3}\left(5 x^{3}-1ight)+8_{C_{4}}} \ {\left(5 x^{3}+1ight)^{2}\left(5 x^{3}-1ight)^{2}+} \ {^{8} C_{6}\left(5 x^{3}+1ight)\left(5 x^{3}-1ight)^{3}+8_{C_{4}}\left(5 x^{3}-1ight)^{4}}\end{array}ight]$

So, the degree of polynomial is $12$ , Now, coefficient of $x^{12}=\left[^{8} \mathrm{C}_{0} 5^{4}+^{8} \mathrm{C}_{2} 5^{4}+^{8} \mathrm{C}_{4} 5^{4}+^{8} \mathrm{C}_{6} 5^{4}+^{8} \mathrm{C}_{8} \mathrm{5}^{4}ight]$

$=5^{4} 	imes \frac{2^{8}}{2}=5^{4} 	imes 2^{4} 	imes \frac{2^{2}}{2}$

$=10^{4} 	imes 2^{3}=8(10)^{4}$

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