Question

P=5-x,prove thatx^3+15px+p^3-125=0.

Answer 1

$Given, \: p \: = 5 - x \\ \\ To \: prove = > {x}^{3} + 15px + {p}^{3} - 125 = 0 \\ \\ Put \: the \: given \: value \: of \: p \: in \: the \: equation \\ \\ = > {x}^{3} + 15(5 - x)x + (5 - {x)}^{3} - 125 = 0 \\ \\ Multiply \: 15x \: with \: 5 - x \: in \: order \: to \: remove \: parenthesis \: \\ \\ {x}^{3} + 15x \times 5 - 15x \times x + (5 - x {)}^{3} - 125 = 0$

${x}^{3} + 75x - 15 {x}^{2} + (5 - x {)}^{3} - 125 = 0 \\ \\ Expand \: (5 - x {)}^{3} \: by \: using \: the \: identitiy \: - - \\ \\ (a - {b)}^{3} = {a}^{3} - 3 {a}^{2} b + 3a {b}^{2} - {b}^{3} \\ \\ {x}^{3} + 75x - 15 {x}^{2} + {5}^{3} - 3 \times {5}^{2} x + 3 \times {5x}^{2} - {x}^{3} - 125 = 0 \\ \\Evaluate \: the \: power \: and \: Calculate \: the \: product \\ \\ {x}^{3} + 75x - 15 {x}^{2} + 125 - 3 \times 25x + {15x}^{2} - {x}^{3} - 125 = 0$

${x}^{3} + 75x - {15x}^{2} + 125 - 75x + {15x}^{2} - {x}^{3} - 125 = 0 \\ \\ Eliminate \: the \: opposites \: and \: sum \: of \: two \: opposites \: equals \: to \: zero \\ \\ Here, \: opposites \: are \: = > \\ \\ {x}^{3} - {x}^{3} = 0 \\ \\ 75x - 75x = 0 \\ \\ - 15 {x}^{2} + 15 {x}^{2} = 0 \\ \\ 125 - 125 = 0 \\ \\ = > 0 + 0 + 0 + 0 = 0 \\ \\ Therefore \\ \\ 0 = 0 \\ \\ As \: L.h.s = R.h.s, \: Hence \: proved$

Answer 2

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