# Math Help - help me out this root problems

1. ## help me out this root problems

question:1. ³√[3(³√x-1/³√x)]=2, then ³√[x]+1/³√[x]= what....

question 2 If a=2b and b =4c, then ³√[a²/16bc]

question 3 the sides of a triangle are denoted by x,y,z. Area of the triangle and semi-perimeter of the triangle are denoted by P and q respectively. If P =√[q(q-x)(q-y)(q-z) and x+y-z = y+z-x = z+x-y = 4. Find P [in square units]

question 4. If a is any natural number, then a³-1/a³ will always be greater than or equal to
option 1. 3{a+1/a} option 2. a+1/a option 3. [a³+1/a³] option 4. 3(a+1/a)

i am only 8th... so please explain clearly.. i already posted the question.. i didn't get the answer and so i again posted.. this time help me out..please..

thanks a lot...

2. ## Re: help me out this root problems

Hi Kohila,

Here are some thoughts on Question 2: If a=2b and b =4c, then ³√[a²/16bc].

*It requires to determine the value of ³√[a²/16bc]. Since the exact values of a, b, c are not provided, we need to know the relationships among a, b, c and make appropriate substitutions to factorize ³√[a²/16bc].
*The provided constraints, i.e., a=2b and b=4c, are just what we want to know---the relationships among a, b, c.
*From a=2b and b=4c we observe that both a and c relate to b. So we could substitute a and c in ³√[a²/16bc] with b. More specifically, we have a=2b and c=b/4.
*Now transform ³√[a²/16bc] to ³√[(2b)²/16b(b/4)] and I'm sure you could proceed with the rest

3. ## Re: help me out this root problems

Hi Kohila,

Here are some thoughts on Question 3. Hope they are helpful.

*the key point to solve p is to determine the values of q, q-x, q-y, q-z.
*semi-perimeter q=(x+y+z)/2, then we have:
q-x=(y+z-x)/2 where (y+z-x)'s value is assigned. So we can determine the value of q-x and that of q-y, q-z similarly.
*Now only the value of q=(x+y+z)/2 is left to determine. Please observe the provided constraints as follows:
x + y - z =4
-x + y + z =4
x - y +z =4
Any clue? All three variables x, y, and z happen three times in three provided equations: two times of + and one time of -. Combine all them then we get exactly (x+y+z) and therefore the value of q.

Now you could try find the values of q, q-x, q-y, q-z accordingly~

4. ## Re: help me out this root problems

Hello, kohila!

$[1]\;\text{Given: }\;\sqrt[3]{3\left(\sqrt[3]{x} - \frac{1}{\sqrt[3]{x}}\right)} \:=\:2$

. . . $\text{Find: }\:\sqrt[3]{x} + \frac{1}{\sqrt[3]{x}}$

Cube the equation: . $3\left(\sqrt[3]{x} - \frac{1}{\sqrt[3]{x}}\right) \;=\;8 \quad\Rightarrow\quad \sqrt[3]{x} - \frac{1}{\sqrt[3]{x}} \;=\;\frac{8}{3}$

Square both sides:

. . . $\left(\sqrt[3]{x} - \frac{1}{\sqrt[3]{x}}\right)^2 \;=\;\left(\frac{8}{3}\right)^2$

. . $\sqrt[3]{x^2} - 2 + \frac{1}{\sqrt[3]{x^2}} \;=\;\frac{64}{9}$

. . . . . $\sqrt[3]{x^2} + \frac{1}{\sqrt[3]{x^2}} \;=\;\frac{82}{9}$

. . $\sqrt[3]{x^2} + 2 + \frac{1}{\sqrt[3]{x^2}} \;=\;\frac{82}{9} + 2$

. . . $\left(\sqrt[3]{x} + \frac{1}{\sqrt[3]{x}}\right)^2 \;=\;\frac{100}{9}$

Therefore: . $\sqrt[3]{x} + \frac{1}{\sqrt[3]{x}} \;=\;\frac{10}{3}$

5. ## Re: help me out this root problems

Originally Posted by Soroban
Hello, kohila!

Cube the equation: . $3\left(\sqrt[3]{x} - \frac{1}{\sqrt[3]{x}}\right) \;=\;8 \quad\Rightarrow\quad \sqrt[3]{x} - \frac{1}{\sqrt[3]{x}} \;=\;\frac{8}{3}$

Square both sides:

. . . $\left(\sqrt[3]{x} - \frac{1}{\sqrt[3]{x}}\right)^2 \;=\;\left(\frac{8}{3}\right)^2$

. . $\sqrt[3]{x^2} - 2 + \frac{1}{\sqrt[3]{x^2}} \;=\;\frac{64}{9}$

. . . . . $\sqrt[3]{x^2} + \frac{1}{\sqrt[3]{x^2}} \;=\;\frac{82}{9}$

. . $\sqrt[3]{x^2} + 2 + \frac{1}{\sqrt[3]{x^2}} \;=\;\frac{82}{9} + 2$

. . . $\left(\sqrt[3]{x} + \frac{1}{\sqrt[3]{x}}\right)^2 \;=\;\frac{100}{9}$

Therefore: . $\sqrt[3]{x} + \frac{1}{\sqrt[3]{x}} \;=\;\frac{10}{3}$
wow, it's really a neat solution!

6. ## Re: help me out this root problems

Thanks a lot...