# Some sums

1. Nov 19, 2009

### 1/2"

Hi there!!
Here are a few sums that are making me go nuts( actually can't get any clue how to solve)
so here they are
( a+x)^1/3 + (a-x)^1/3= b
( Gosh I wish there could be some rule so that we could straight away write a^ 1/3 and x^1/3 anyway:tongue:)
And 1/p+q+x=1/p+1/q+1/x
I WOULD BE RALLY VERY THANKFUL IF ANYONE COULD GIVE A "HINT"!!
THANK YOU

Last edited: Nov 19, 2009
2. Nov 19, 2009

### Staff: Mentor

When you have fractional exponents written as you have them, use parentheses around the exponent. The 1/3 power is the same as the cube root.

$$\sqrt[3]{a + x} + \sqrt[3]{a + x}~=~2\sqrt[3]{a + x}$$
Now, divide both sides of your equation by 2, and then cube both sides.

For you other problem, 1/p+q+x=1/p+1/q+1/x, it's hard to say anything without knowing exactly what the problem is -- use parentheses.
As you have written it, the problem is (1/p) + q + x = 1/p+1/q+1/x. I suspect that you meant the left side to be 1/(p + q + x), but I'm not sure.

Last edited: Nov 19, 2009
3. Nov 19, 2009

### 1/2"

Ya it's 1/(p+q+x)

4. Nov 19, 2009

### 1/2"

Sorry i got the sign wrong it should be
( a+x)^1/3 + (a-x)^1/3= b
I am REALLY VERY SORRRRRY!!!

5. Nov 19, 2009

### Staff: Mentor

That makes it a different, and harder, problem...

6. Nov 19, 2009

### Count Iblis

Substitute x =a + t^3 to write this as:

(2a + t^3)^1/3 - t = b ------->

(2a + t^3)^1/3 = b + t

Take the cube of both sides. You'll see that the t^3 term cancels, so you have a quadratic equation for t.

7. Nov 22, 2009

### 1/2"

Thanks A LOT .
I have figured the answer !!