Math, asked by lincyjoseph281092, 1 month ago

If p²-3p-1=0,find
iii)p³-1/p³

Answers

Answered by ranii410
0

Answer:

p² - 3p + 1 = 0

_____________ [GIVEN]

• We have to find the value of p² + \dfrac{1}{ {p}^{2} }

p

2

1

______________________________

Solution:

\implies⟹ p² - 3p + 1 = 0

\implies⟹ p² + 1 = 3p

Divide by p on both sides

\implies⟹ \dfrac{ {p}^{2} }{p}

p

p

2

+ \dfrac{1}{p}

p

1

= \dfrac{3p}{p}

p

3p

\implies⟹ p + \dfrac{1}{p}

p

1

= 3

Now.. squaring on both sides..

\implies⟹ \bigg ({p \: + \: \dfrac{1}{p}} \bigg)^{2}(p+

p

1

)

2

= (3)²

• (a + b)² = a² + b² + 2ab

\implies⟹ \bigg ( {p}^{2} \: + \: \dfrac{1}{ {p}^{2} } \: + \: 2p \: \times \: \dfrac{1}{p} \bigg)(p

2

+

p

2

1

+2p×

p

1

) = 9

\implies⟹ p² + \dfrac{1}{ {p}^{2} }

p

2

1

+ 2 = 9

\implies⟹ p² + \dfrac{1}{ {p}^{2} }

p

2

1

= 9 - 7

_____________________________

\huge{\bold{{p}{^2}\:+\:\dfrac{1}{p}^{2} \: =\:7}}p

2

+

p

1

2

=7

Answered by mohanspandit001
0

Step-by-step explanation:

p² - 3p + 1 = 0

_____________ [GIVEN]

• We have to find the value of p² + \dfrac{1}{ {p}^{2} }p21

______________________________

Solution:

\implies⟹ p² - 3p + 1 = 0

\implies⟹ p² + 1 = 3p

Divide by p on both sides

\implies⟹ \dfrac{ {p}^{2} }{p}pp2 + \dfrac{1}{p}p1 = \dfrac{3p}{p}p3p

\implies⟹ p + \dfrac{1}{p}p1 = 3

Now.. squaring on both sides..

\implies⟹ \bigg ({p \: + \: \dfrac{1}{p}} \bigg)^{2}(p+p1)2 = (3)²

• (a + b)² = a² + b² + 2ab

\implies⟹ \bigg ( {p}^{2} \: + \: \dfrac{1}{ {p}^{2} } \: + \: 2p \: \times \: \dfrac{1}{p} \bigg)(p2+p21+2p×p1) = 9

\implies⟹ p² + \dfrac{1}{ {p}^{2} }p21 + 2 = 9

\implies⟹ p² + \dfrac{1}{ {p}^{2} }p21 = 9 - 7

_____________________________

\huge{\bold{{p}{^2}\:+\:\dfrac{1}{p}^{2} \: =\:7}}p2+p12=7

_____________ \bold{[ANSWER]}[ANSWER]

_____________________________

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