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Q. (b) The value of the constant term in the trinomial $\left(x^2+\frac{1}{x^2}-2\right)^{10}$ is also equal to

Binomial Theorem

Solution:

$\left(x^2+\frac{1}{x^2}-2\right)^{10}=\left(x-\frac{1}{x}\right)^{20}$
$T _{ r +1}={ }^{20} C _{ r } x ^{20- r }(-1)^{ r } \frac{1}{ x ^{ r }}={ }^{20} C _{ r } x ^{20-2 r }(-1)^{ r } $
$\therefore 20-2 r =0 \Rightarrow r =10$
So, term independent of $x ={ }^{20} C _{10}$. Now verify alternatives