find the degree of the following expression:
(√2x2+1 + √2x2-1)6+ 26[√2x2+1 + √2x2-1 ]-6
1√2x2+1 = x[√2+(1/x2)]
Do similarly 4 the other term
This when raised to 6 gives x6[√{2+(1/x2)}+√{2-(1/x2)}]6
The [√{2+(1/x2)}+√{2-(1/x2)}]6 wala part part will giv U a constt. on expansion
Ise x6 se multiply karne pe U will get a term of power 6
The second term when expanded will have terms of power less than 6
So the degree of the expression is 6
1sorry i could nt get how did you get that the expansion of the second term will have powers less than 6
62[√2x2+1 + √2x2-1 ]
can be written as (\sqrt{2x^2+1}+\sqrt{2x^2-1})\frac{\sqrt{2x^2+1}-\sqrt{2x^2-1}}{\sqrt{2x^2+1}-\sqrt{2x^2-1}}
=\frac{2}{\sqrt{2x^2+1}-\sqrt{2x^2-1}}
Hence by replacing the second term the whole expression becomes
[√2x2+1 + √2x2-1 ]6+[√2x2+1 - √2x2-1 ]6
try with this in mind..