Question

If p+1/p=6 find 3p cube +3/p cube

Answer 1

Given : p + 1/p = 6

To find : 3p^3 + 3/p^3

3 (p^3 + 1/p^3)

$( {x}^{3} + {y}^{3} )=(x+y)( {x}^{2} + {y}^{2} −xy)$

3 (p + 1/p) (p^2 + 1/p^2 - p×1/p)

3 × 6 × (p^2 + 1/p^2 - 1)

$({x + y})^{2} = {x}^{2} + {y}^{2} + 2xy$

so, x^2 + y^2 = (x + y)^2 - 2xy

18 × [ (p + 1/p)^2 - 2p×1/p -1 ]

18 × [ 6^2 - 2 - 1 ]

18 × [ 36 - 3 ]

18 × 33

= 594