Question
If ω is the complex cube root of unity, show that
$(a+b)^2+\left(a \omega+b \omega^2\right)^2+\left(a \omega^2+b \omega\right)^2=6 a b$
$(a+b)^2+\left(a \omega+b \omega^2\right)^2+\left(a \omega^2+b \omega\right)^2=6 a b$
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$\frac{n !}{3 !(n-3) !}: \frac{n !}{5 !(n-5) !}=5: 3$
(i) the first card drawn is kept aside.
ii. the first card drawn is replaced in the pack.