Question

answer question with photos​

Answer 1

Answer:

Heya here's your answer 5

Step-by-step explanation:

In the attachment

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Answer 2

Given $x^3+\dfrac{1}{x^3}=110$

Recall that $\bigg(x+\dfrac{1}{x}\bigg)^3=x^3+\dfrac{1}{x^3}+3\times\bigg(x+\dfrac{1}{x}\bigg)$

Let $\bigg(x+\dfrac{1}{x}\bigg)=p$

$\implies p^3 = 110 + 3p$

$\implies p^3-3p-110=0$

Put $p=5$

$\implies 5^3-3\cdot(5)-110$

$=125-15-110=0$

Hence $(p - 5)$ is a factor.

On dividing $(p^3-3p-110)$ with $(p-5)$ we get the quotient as $(p^2+5p+22)$

$\therefore p^3-3p-110=(p-5)(p^2+5p+22) = 0$

Hence $(p-5)=0$ and $(p^2+5p+22)\neq0$

$\implies\bigg(x+\dfrac{1}{x}\bigg)-5=0$

$\therefore \bold{\bigg(x+\dfrac{1}{x}\bigg)=5}$