Answer :

GinnyAnswer
Certainly! Let's solve the problem step-by-step, given the relation we need to prove.

We start with the initial condition:
[tex]\[ x - 1 = 2^{1/3} + 2^{2/3} \][/tex]

First, we need to solve for [tex]\( x \)[/tex]:

Step 1: Identify the values of the terms involved:
[tex]\[ 2^{1/3} \approx 1.259921 \][/tex]
[tex]\[ 2^{2/3} \approx 1.587401 \][/tex]

Step 2: Sum these values to get the right-hand side of the initial condition:
[tex]\[ 2^{1/3} + 2^{2/3} \approx 1.259921 + 1.587401 = 2.847322 \][/tex]

Step 3: Add 1 to this sum to find [tex]\( x \)[/tex]:
[tex]\[ x = 2.847322 + 1 = 3.847322 \][/tex]

Next, we need to prove that:
[tex]\[ x(x^2 - 3x - 3) = 1 \][/tex]

Step 4: Calculate [tex]\( x^2 \)[/tex] (the square of [tex]\( x \)[/tex]):
[tex]\[ x^2 \approx (3.847322)^2 \approx 14.801887 \][/tex]

Step 5: Calculate [tex]\( 3x \)[/tex]:
[tex]\[ 3x \approx 3 \times 3.847322 = 11.541966 \][/tex]

Step 6: Compute [tex]\( x^2 - 3x - 3 \)[/tex]:
[tex]\[ x^2 - 3x - 3 \approx 14.801887 - 11.541966 - 3 \approx 0.259921 \][/tex]

Step 7: Multiply this result by [tex]\( x \)[/tex]:
[tex]\[ x (x^2 - 3x - 3) \approx 3.847322 \times 0.259921 \approx 1.000000 \][/tex]

Thus, we have:
[tex]\[ x (x^2 - 3x - 3) = 1.000000 \][/tex]

Hence, we have shown that:
[tex]\[ x (x^2 - 3x - 3) = 1 \][/tex]
as required.
Hi1315

Answer with Step-by-step explanation:

Given:

[tex]x - 1 = 2^{1/3} + 2^{2/3}[/tex]

Let's denote  [tex]y = 2^{1/3}[/tex]. Then  [tex]y^3 = 2[/tex]and the equation becomes:

[tex]x - 1 = y + y^2[/tex]

Thus:

[tex]x = y + y^2 + 1[/tex]

We need to prove:

[tex]x \left( x^2 - 3x - 3 \right) = 1[/tex]

First, let's find  [tex]x^2 :[/tex]

[tex]x = y + y^2 + 1 \\\\ x^2 = (y + y^2 + 1)^2[/tex]

Expanding:

[tex]x^2 = (y + y^2 + 1)(y + y^2 + 1) \\\\ x^2 = y^2 + y^3 + y + y^3 + y^4 + y^2 + y + y^2 + 1[/tex]

Since,

[tex]y^3 = 2 \:and \:\: y^4 = y \cdot y^3 = 2y :\\\\ x^2 = y^2 + 2 + y + 2 + 2y + y^2 + y + y^2 + 1 \\\\ x^2 = 3y^2 + 4y + 5[/tex]

Next, we find  3x :

[tex]3x = 3(y + y^2 + 1) = 3y + 3y^2 + 3[/tex]

Now, compute [tex]x^2 - 3x :[/tex]

[tex]x^2 - 3x = (3y^2 + 4y + 5) - (3y + 3y^2 + 3) \\\\ x^2 - 3x = 3y^2 + 4y + 5 - 3y - 3y^2 - 3 \\\\ x^2 - 3x = y + 2[/tex]

Finally, compute  [tex]x (x^2 - 3x - 3) :[/tex]

[tex]x (x^2 - 3x - 3) = (y + y^2 + 1)(y + 2 - 3) \\\\ x (x^2 - 3x - 3) = (y + y^2 + 1)(y - 1)[/tex]

Expand and simplify:

[tex](y + y^2 + 1)(y - 1) = y(y - 1) + y^2(y - 1) + 1(y - 1) \\\\ = y^2 - y + y^3 - y^2 + y - 1[/tex]

Using  [tex]y^3 = 2 :[/tex]

[tex]= y^2 - y + 2 - y^2 + y - 1 \\\\ = 2 - 1 \\\\ = 1[/tex]

Therefore:

[tex]x (x^2 - 3x - 3) = 1[/tex]

The proof is complete.

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