Question 119

If 2p + $$\frac{1}{p}$$ = 4 the value of p3 + $$\frac{1}{8p^3}$$

Solution

Given 2p + $$\frac{1}{p}$$ = 4
or p + $$\frac{1}{2p}$$ = 2
taking cube on both sides we will get
$$(p^{3} + \frac{1}{8p^{3}}) + 3( \frac{p}{2} + \frac{1}{4p} )$$ = 8
as $$( \frac{p}{2} + \frac{1}{4p} ) $$ will be equal to 1 from the given eq.
hence $$(p^{3} + \frac{1}{8p^{3}})$$ = 5


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