Question

Solve pls, ans should be 0, add working

Answer 1

Step-by-step explanation:

Given: {x+(1/x)}³ = 3

Asked: x³ + (1/x³) = ?

Solution:

Method 1:

We have, {x+(1/x)}³ = 3

Comparing the expression with (a+b)³, we get

a = x

b = (1/x)

Using identity (a+b)³ = a³+b³+3ab(a+b), we get

⇛{x+(1/x)}³ = 3

⇛(x)³ + (1/x)³ + 3(x)(1/x){x + (1/x)} = 3

⇛(x*x*x) + (1*1*1/3*3*3) + 3(x)(1/x){x + (1/x)} = 3

⇛x³ + (1/x³) + 3(x)(1/x){x + (1/x)} = 3

⇛x³ + (1/x³) + 3{x + (1/x)} = 3

⇛x³ + (1/x³) + 3(x) + 3(1/x) = 3

⇛x³ + (1/x³) + 3x + (3/x) = 3

Our answer came incorrect.

Let's try..

Method 2:

We have,

[x+(1/x)]³ = 3

On taking cube root both sides then

⇛³√[{ x+(1/x)}³ ] = ³√3

⇛x+(1/x) = ³√3 -----(1)

We know that

a³+b³ = (a+b)³-3ab(a+b)

⇛x³+(1/x)³ = [x+(1/x)]³ - 3(x)(1/x)[x+(1/x)]

⇛x³+(1/x³) = (3)-3(1)(³√3)

[since, {x + (1/x)} = ³√3 from equation (1)]

⇛x³+(1/x)³ = 3-3 ׳√3

⇛x³ + (1/x³) = 3- ³√81 (or )

⇛x³ + (1/x³) = 3(1-³√3)

Therefore, x³ + (1/x³) = 3(1 - cube root of 3)

It is impossible to get zero

Answer 2

Based on the calculations, the expression $x^3 +(\frac{1}{x})^3$ is equal to $3(1-\sqrt[3]{3})$

Given the following data:

• $(x + \frac{1}{x} )^3=3$

• $x^3 +\frac{1}{x^3}$

How to solve the equation.

First of all, we would take the cube root of both sides as follows:

$\sqrt[3]{(x + \frac{1}{x} )^3} =\sqrt[3]{3} \\\\x + \frac{1}{x} =\sqrt[3]{3}$....equation 1.

From trinomial, we have:

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

Applying the trinomial eqn. & substituting eqn. 1, we have:

$x^3 +(\frac{1}{x})^3 = [x+\frac{1}{x}]^3 - 3(x)(\frac{1}{x})[x+\frac{1}{x}]\\\\x^3 +(\frac{1}{x})^3 = (\sqrt[3]{3})^3 - 3[x+\frac{1}{x}]\\\\x^3 +(\frac{1}{x})^3 =3-3\sqrt[3]{3} \\\\x^3 +(\frac{1}{x})^3 =3(1-\sqrt[3]{3})$

Read more on trinomial here: https://brainly.com/question/11777906