Given the sum x + 1/x = 4 what is x^2 + 1/x^2, x^3 + 1/x^3 ?

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hala718's profile pic

hala718 | High School Teacher | (Level 1) Educator Emeritus

Posted on

x+ 1/x = 4

Let us square both sides:

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

==? x^2 + 2 + 1/x^2 = 16

Reaarange terms:

==> x^2 ++ 1/x^2 + 2 = 16

Now subtract 2 from both sides:

==> x^2 + 1/x^2 = 14

 

x^3 + 1/x^3 = ?

x+ 1/x = 4

Now cube both sides:

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

==> (x^2 + 1/x^2 + 2) (x+ 1/x) = 64

= (x^3 + x + 1/x + 1/x^3 + 2x + 2/x = 64

Rearrange sides:

==> (x^3 + 1/x^3 + x + 1/x + 2x + 2/x) = 64

let us factor 2 :

==> x^3 + 1/x^3 + x+ 1/x + 2(x+ 1/x) = 64

but we know that x+ 1/x = 4

==? x^3 + 1/x^3 + 4 + 2*4 = 64

==> x^3 + 1/x^3 + 12 = 64

Now subtract 12 from both sides:

==> x^3 + 1/x^3 = 52

giorgiana1976's profile pic

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

To calculate the first sum:

x^2 + 1/x^2

we'll have to raise to square the given sum:

x + 1/x = 4

We'll raise to square both sides:

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

We'll apply the formula of the binomial raised to square:

(a+b)^2 = a^2 + 2ab + b^2

x^2 + 2*x*(1/x) + (1/x)^2 = 16

We'll hold to the left side only the sum of the squares:

x^2 + (1/x)^2 = 16 - 2*x*(1/x)

We'll simplify and we'll get:

x^2 + (1/x)^2 = 16 - 2

x^2 + (1/x)^2 = 14

To calculate the sum of the cubes, we'll apply the formula:

a^3 + b^3 = (a+b)(a^2 - ab + b^2)

x^3 + 1/x^3 = (x + 1/x)(x^2 - x*1/x + 1/x^2)

We'll simplify and we'll substitute x^2 + 1/x^2 by the result 14:

x^3 + 1/x^3 = 4*(14 - 1)

x^3 + 1/x^3 = 4*13

x^3 + 1/x^3 = 52

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