P for Practice

$p^3 + \frac{1}{p^3} = p^3 + (\frac{1}{p})^3 \Longrightarrow$
$p^3 + \frac{1}{p^3} = (p + \frac{1}{p})(p^2 - p\frac{1}{p} + \frac{1}{p^2}) \Longrightarrow$
$p^3 + \frac{1}{p^3} = (p + \frac{1}{p})(p^2 + \frac{1}{p^2} - p\frac{1}{p}) \Longrightarrow$
$p^3 + \frac{1}{p^3} = (p + \frac{1}{p})(1 - 1) \Longrightarrow$
$p^3 + \frac{1}{p^3} = (p + \frac{1}{p}) \times 0 \Longrightarrow$
$p^3 + \frac{1}{p^3} = \boxed{0}$

Since we know that

$\left(p+\frac{1}{p}\right)^2=p^2+\frac{1}{p^2}+2=3$

$\implies p+\frac{1}{p}=\pm \sqrt{3}$

And also we know

$\left(p+\frac{1}{p}\right)^3=p^3+\frac{1}{p^3}+3\left(p+\frac{1}{p}\right)$

$\implies p^3+\frac{1}{p^3}=\left(p+\frac{1}{p}\right)^3-3\left(p+\frac{1}{p}\right)$

$\implies p^3+\frac{1}{p^3}=\pm 3.\sqrt{3} \mp 3.\sqrt{3}=\boxed{0}$

enjoy !

Even though there is a proven answer to this answer. But from some logical tests, it seems like there cannot exist a real number p that give the equation (p^2)+1/(p^2)=1.

If p=1, then (p^2)+1/(p^2)=2. If p equals any number higher or lower than 1, then (p^2)+1/(p^2)>2.

Multiply both sides by $p+\frac{1}{p}$ to get:

$p^3+\frac{1}{p^3}+p+\frac{1}{p}=p+\frac{1}{p}$

$p^3+\frac{1}{p^3}=0$

We know that $p+\frac{1}{p}$ is not 0 because that is only possible if $p$ is purely complex (which it isn't, because then $p^2+\frac{1}{p^2}$ has no solutions). Hence why we can multiply both sides by it.

We can transform the original equation by multiplying through by $p^2$ (keeping in mind that $p=0$ cannot be a solution), then bringing all the terms on one side.

$p^4 - p^2 + 1 = 0$

We can use the quadratic formula to solve for $p^2$ :

$p^2 = \frac{1 \pm \sqrt{-3}}{2}$

$p^2 = \frac{1}{2} \pm \frac{\sqrt{3}}{2} i$

This can easily be transformed into an exponent using Euler's equation:

$p^2 = \exp\left(\pm i \pi/3\right)$

Taking the square root of both sides, we get:

$p = \pm \exp\left(\pm i \pi/6\right)$

Our desired expression involves $p^3$ , so we raise both sides to the power 3:

$p^3 = \pm \exp\left(\pm i \pi/2\right)$

Using Euler's equation again, we can transform this to:

$p^3 = \pm i$

Plugging into the final expression:

$p^3 + \frac{1}{p^3}$

$\pm i + \frac{1}{\pm i}$

$\pm i \mp i = 0$

Thus the expression evaluates to 0.

Multiply p to get :: p^3 = p - 1/p Multiply with 1/p :: 1/P^3 = 1/p - p

Therefore, p^3 + 1/p^3 = 0

p^2 + 1/p^2 = 1

p^4 - p^2 + 1 = 0

x = p^2

x^2 - x + 1 = 0

x1 = 1/2 - 1/2sqrt(3)i

x2 = 1/2 + 1/2sqrt(3)i

p1 = -sqrt(1/2 - 1/2sqrt(3)i)

p2 = sqrt(1/2 - 1/2sqrt(3)i)

p3 = -sqrt(1/2 + 1/2sqrt(3)i)

p4 = sqrt(1/2 + 1/2sqrt(3)i

p^3 + 1/p^3 = sqrt(1/2 + 1/2sqrt(3)i)^3 + 1/sqrt(1/2 + 1/2sqrt(3)i)^3 (any p can be used)

(1/2 + 1/2sqrt(3)i)^3

= 1/8 + 3/8sqrt(3)i - 9/8 - 3/8sqrt(3)i

= -1

p^3 + 1/p^3

= sqrt(-1) + 1/sqrt(-1)

= i + 1/i

= i - i

= 0

$p^2+\frac{1}{p^2}=1 \Rightarrow \frac{p^4+1}{p^2}=1 \Rightarrow p^4-p^2+1=0$ Now since $p^2+\frac{1}{p^2}=1 \Rightarrow p^3+\frac{1}{p}=p \Rightarrow p^3=\frac{p^2-1}{p}$ Therefore $p^3+\frac{1}{p^3}=\frac{p^2-1}{p}+\frac{p}{p^2-1}=\frac{p^4-2p^2+1+p^2}{p(p^2-1)}=\frac{p^4-p^2+1}{p(p^2-1)}=0$ We already concluded numerator is zero

if we multiple $P^2+1/P^2=1$ with $p \\and\\ 1/p$ then we wil got $P^3=P-(1/P) \\and\\ 1/P^3=(1/P)-P$ hence the answer is 0

p2 + 1/p2 =1 >> (p +1/p)2 -2 = 1>> Establishing (p +1/p) = 3.

Also, we know that p3 +1/p3 = (p +1/p)^3 - 3( p+1/p) ; Hence , substituting the value of p + 1/p.

use whole square $2^{(a-b)} = 2^{a} + 2^{b} - 2ab$ you will get a-b as $\sqrt{3}$ now we find a-b value now cube it take the 3ab(a-b) on other side which will be same as the whole cube $\sqrt{3}$ subtracting gets 0

(p+1/p)^2 -2.(p 1/p)=1 p+1/p=sq. root 3 again from the question we can write , (p+1/p)^3 -3 p*1/p (p+1/p) then, 3( sq root 3 ) - 3( sq root 3)

>>> 0

that's the answer !!!!!!!!!!!!!!!!!!!

{ p }^{ 3 }+\frac { 1 }{ { p }^{ 3 } } \ =\quad \frac { ({ p }^{ 6 }+1) }{ { p }^{ 3 } } \ =\quad \frac { ({ p }^{ 2 }+1)({ p }^{ 4 }-{ p }^{ 2 }+1) }{ { p }^{ 3 } } \ Since\quad { p }^{ 2 }+\frac { 1 }{ { p }^{ 2 } } =\quad 1,\quad multiplying\quad by\quad { p }^{ 2 }\quad yields\quad { p }^{ 4 }-{ p }^{ 2 }+1=0,\quad which\quad makes\quad the\quad above\quad expression\quad equal\quad to\quad 0.

p2+1/p2+2.p.1/p=1+2 (p+1/p)2=3 multiply with (p+1/p) on both sides (p+1/p)3=p3+3.p2.1/p+3.p.1/p2+1/p3 =>p3+1/p3+3(p+1/p)=3(p+1/p) =>p3+1/p3=0

p=±cis(±PI/6) are 4 solutions of first equation,so p^3=±cis(±PI/2),p^(-3)=±cis(∓PI/2) so p^3+p^(-3)=±cis(±PI/2)±cis(∓PI/2)=±(cos(PI/2)±isin(PI/2))±(cos(PI/2)∓isin(PI/2))=±(±i)±(∓i)=0 for all 4 solutions of p...