Question

if (2x+1/X) = 3 , then find the value of (8x³ + 1/x³).​

Answer 1

Answer:

If (2x)+ (1/2x) =3 then (x³)+ (1/8x³)=?

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I am not really sure if someone indeed needs any help given the way the question is formatted here, this is a nagging bother. Nevertheless, I will solve both, what the question is given and what the question might have been.

Case 1: If 2x+12x=3 then x3+18x3=? (just rewritten whatever is given, in more legible format)

This is indeed very simple as: 2x+12x=2.5x=3

Thus, x=32.5=1.2=65

Therefore, x3+18x3=[1+18]x3=[1+18](65)3=243125

Case 2: If 2x+12x=3 then 8

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Given: (2x)+(1/2x)=3

(4x^2+1)/2x=3

4x^2+1=6x

4x^2-6x+1=0

By formula method

x=(3+root5)/4 and (3-root5)/4

And we know that :root 5=2.23 (approx)

So,

x=1.3 and 0.2

Now,

(x^3)+(1/8x^3)=?

First we put x=1.3 then we get

=2.15

And when we put x=0.2 then we get

=15.6

Step-by-step explanation:

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Answer 2

Answer:

$\huge\bf\underline\green{AnSweR:}$

$( {2x + \frac{1}{x} })^{3} = ( {2x})^{3} + ({ \frac{1}{x} })^{3} + 3(2x)$

$( \frac{1}{x} )(2x + \frac{1}{x})$

$= ( {3})^{3}$

${8x}^{3} + \frac{1}{ {x}^{3} } + 6(2x + \frac{1}{x} ) = 27$

${8x}^{3} + \frac{1}{ {x}^{3} } + 6(3) = 27$

${8x}^{3} + \frac{1}{ {x}^{3} } = 27 - 18$

$\implies\bf\red{9}$