Question

p+1/p=4 ,prove that p³+1/p³=52​

Answer 1

Answer:

Proof is given below.

Step-by-step explanation:

Given that

$p+\frac{1}{p} =4$

Cubing on both sides,

$(p+\frac{1}{p})^{3}=64$

$p^{3} +\frac1{p^3}$ $+3(p)(\frac{1}{p})(p+\frac{1}{p} )$ = 64

But, we know that p × $\frac{1}{p}$ = 1 and $p+\frac{1}{p} =4$. So this gives us

$p^3+ \frac{1}{p^3} + 3(1)(4) = 64$

⇒ $p^3+ \frac{1}{p^3} + 12 = 64$

⇒$p^3+\frac{1}{p^3} = 52$

Hence proved.

Answer 2

Answer:

GIVEN :-

$\sf \: p + \frac{1}{p} = 4$

TO PROVE :-

$\sf \: p^{3} + \frac{1}{p^{3} } = 52$

PROVING :-

$\sf \: It \: \: is \: \: given \: \: that$

$\sf \: p + \frac{1}{p} = 4$

$\sf \: Taking \: \: cube \: \: on \: \: \: \: both \: \: sides$

$\implies \sf \bigg( p + \frac{1}{p} \bigg)^{3} = {4}^{3}$

$\sf \: Formula \: -$

$\orange{\sf \: (x + y) {}^{3} = {x}^{3} + {y}^{3} + 3xy(x + y)}$

$\sf \: By \: \: using \: \: the \: \: formula \: \: we \: \: get -$

$\implies \sf {p }^{3} + \frac{1}{ {p}^{3} } + 3(p).\frac{1}{(p)} \bigg(p + \frac{1}{p} \bigg) = 64$

$\implies \sf {p }^{3} + \frac{1}{ {p}^{3} } + 3 \bigg(p + \frac{1}{p} \bigg) = 64$

$\sf \: Putting \: \: the \: value \: of \: ( p + \frac{1}{p} ) = 4$

$\implies \sf {p }^{3} + \frac{1}{ {p}^{3} } + 3 \times 4 = 64$

$\implies \sf {p }^{3} + \frac{1}{ {p}^{3} } + 12= 64$

$\implies \sf {p }^{3} + \frac{1}{ {p}^{3} } = 64 - 12$

$\implies \sf {p }^{3} + \frac{1}{ {p}^{3} } = 52$

$\large \sf \green{ \:Proved}$