IMO 2016, class 8 Chapter 5 question 3 Cubes and Cube roots

[prat]

IMO 2016, Chapter 5 question 3

if cuberoot of {3 * [ cuberootof(x) - 1/cuberootof(x)]} =2
then
cuberoot(x) + 1/cuberoot(x) = ???

---

\(\displaystyle \mbox{3. If }\, \sqrt[3]{3\left(\sqrt[3]{x\,}\, -\, \dfrac{1}{\sqrt[3]{x\,}}\right)\,}\, =\, 2\)

. . .\(\displaystyle \mbox{then find the value of }\, \sqrt[3]{\strut x\,}\, +\, \dfrac{1}{\sqrt[3]{\strut x\,}}\)

 

[Bob Brown MSEE]

8/3, do you see why?

 

[Deleted member 4993]

IMO 2016, Chapter 5 question 3

if cuberoot of {3 * [ cuberootof(x) - 1/cuberootof(x)]} =2
then
cuberoot(x) + 1/cuberoot(x) = ???

---

\(\displaystyle \mbox{3. If }\, \sqrt[3]{3\left(\sqrt[3]{x\,}\, -\, \dfrac{1}{\sqrt[3]{x\,}}\right)\,}\, =\, 2\)

. . .\(\displaystyle \mbox{then find the value of }\, \sqrt[3]{\strut x\,}\, +\, \dfrac{1}{\sqrt[3]{\strut x\,}}\)

If $\displaystyle \displaystyle{ \sqrt[3]{ 3 * \left[\sqrt[3]{x} - \frac{1}{\sqrt[3]{x}}\right]} =2\\ then\\ \sqrt[3]{x} + \frac{1}{\sqrt[3]{x}} =?}$

replace:

y = x^(1/3) then you got

y - (1/y) = 8/3

use:

(a + b)^2 = (a-b)^2 + 4ab

...... continue....

 

[prat]

Thank you

If $\displaystyle \displaystyle{ \sqrt[3]{ 3 * \left[\sqrt[3]{x} - \frac{1}{\sqrt[3]{x}}\right]} =2\\ then\\ \sqrt[3]{x} + \frac{1}{\sqrt[3]{x}} =?}$

replace:

y = x^(1/3) then you got

y - (1/y) = 8/3

use:

(a + b)^2 = (a-b)^2 + 4ab

...... continue....

Thank you very much for this quick and detailed answer.
I was just missing that ab can be ignored here as they both are reciprocal

 

[lookagain]

8/3, do you see why?

That is not the answer. That is what one-third of what the radicand equals.

In other words, that is what cbrt(x) - 1/cbrt(x) equals.


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Anyways, here's another method.

$\displaystyle \displaystyle{ \sqrt[3]{3 \cdot \left[\sqrt[3]{x} - \frac{1}{\sqrt[3]{x}}\right]}} = 2$

If we cube each side, we get:

$\displaystyle \displaystyle{3 \cdot \left[\sqrt[3]{x} - \frac{1}{\sqrt[3]{x}}\right]} = 8$

Dividing each side by 3, we get:

$\displaystyle \displaystyle{\sqrt[3]{x} - \frac{1}{\sqrt[3]{x}}} = \frac{8}{3}$

Again, that is not the answer. It is not answering the question. The question asks for a sum. The radicand is three times a difference.


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If $\displaystyle \displaystyle{ \sqrt[3]{ 3 * \left[\sqrt[3]{x} - \frac{1}{\sqrt[3]{x}}\right]} =2\\ then\\ \sqrt[3]{x} + \frac{1}{\sqrt[3]{x}} =?}$

replace:

y = x^(1/3) then you got

y - (1/y) = 8/3

use:

(a + b)^2 = (a-b)^2 + 4ab

...... continue....

Or, multiply both sides by the LCD, 3y, and get a quadratic:

$\displaystyle 3y^2 - 3 = 8y$

$\displaystyle 3y^2 - 8y - 3 = 0$

$\displaystyle (3y + 1)(y - 3) = 0$

$\displaystyle y = x^{1/3},$

so this equation will have two solutions for x after substituting back.


Conclusion:


$\displaystyle \sqrt[3]{x} + \frac{1}{\sqrt[3]{x}}$ will have two distinct solutions (values). I'll leave it to others to do/attempt the work.

 

[mmm4444bot]

Thank you, lookagain. I have deleted my post.

Grandpa Bob and I need to find our specs.

 

[stapel]

\(\displaystyle \mbox{3. If }\, \sqrt[3]{3\left(\sqrt[3]{x\,}\, -\, \dfrac{1}{\sqrt[3]{x\,}}\right)\,}\, =\, 2\)

. . .\(\displaystyle \mbox{then find the value of }\, \sqrt[3]{\strut x\,}\, +\, \dfrac{1}{\sqrt[3]{\strut x\,}}\)

Following on from scraps of previous posts, we make the substitution:

. . . . .\(\displaystyle y\, =\, \sqrt[3]{\strut x\,}\)

Then the given equation becomes:

. . . . .$\displaystyle \sqrt[3]{3\, \left(y\, -\, \dfrac{1}{y}\right)\,}\, =\, 2$

. . . . .$\displaystyle 3\, \left(y\, -\, \dfrac{1}{y}\right)\, =\, 8$

. . . . .$\displaystyle y\, -\, \dfrac{1}{y}\, =\, \dfrac{8}{3}$

We know that the following are always true:

. . . . .$\displaystyle (a\, -\, b)^2\, =\, a^2\, -\, 2ab\, +\, b^2$

. . . . .$\displaystyle (a\, +\, b)^2\, =\, a^2\, +\, 2ab\, +\, b^2$

In effect, we have:

. . . . .$\displaystyle a\, -\, b$

...and we want:

. . . . .$\displaystyle a\, +\, b$

Subtracting the two identities above gives us:

. . . . .$\displaystyle (a\, +\, b)^2\, -\, (a\, -\, b)^2 = 4ab$

...so:

. . . . .$\displaystyle (a\, +\, b)^2\, =\, (a\, -\, b)^2\, +\, 4ab$

Note that, in our case, we have:

. . . . .$\displaystyle ab\, =\, (y)\left(\dfrac{1}{y}\right)\, =\, 1$

...so the above simplifies as:

. . . . .$\displaystyle (a\, +\, b)^2\, =\, \dfrac{64}{9}\, +\, 4$

Simplify, and take the square root of either side. Since they only asked for the value of the sum, there is no need to attempt to back-solve for x.