If 2p + frac{1}{p} = 4,then the value of p^{3 | vedclass

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(B) Given equation: $2p + \frac{1}{p} = 4$.Divide the entire equation by $2$:$p + \frac{1}{2p} = 2$.Now,cube both sides of the equation:$\left(p + \fr

Vedclass · Accepted Answer

$5$

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(B) Given equation: $2p + \frac{1}{p} = 4$.Divide the entire equation by $2$:$p + \frac{1}{2p} = 2$.Now,cube both sides of the equation:$\left(p + \frac{1}{2p}\right)^{3} = 2^{3}$.Using the identity $(a + b)^{3} = a^{3} + b^{3} + 3ab(a + b)$:$p^{3} + \left(\frac{1}{2p}\right)^{3} + 3(p)\left(\frac{1}{2p}\right)\left(p + \frac{1}{2p}\right) = 8$.$p^{3} + \frac{1}{8p^{3}} + \frac{3}{2}\left(p + \frac{1}{2p}\right) = 8$.Substitute the value $p + \frac{1}{2p} = 2$ into the equation:$p^{3} + \frac{1}{8p^{3}} + \frac{3}{2}(2) = 8$.$p^{3} + \frac{1}{8p^{3}} + 3 = 8$.$p^{3} + \frac{1}{8p^{3}} = 8 - 3 = 5$.

If $2p + \frac{1}{p} = 4$,then the value of $p^{3} + \frac{1}{8p^{3}}$ is:

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