1. ## Find without solving?!

I came across this question in Cambridge. I tried for ages, and I couldn't get it.

For the equation x^2 = 5x - 8, find the value of a^(1/3) + b^(1/3), where a and b are the roots of the equation, without solving the equation.

Thanks.

2. ## Re: Find without solving?!

Why can't you solve the equation?

3. ## Re: Find without solving?!

Originally Posted by Prove It
Why can't you solve the equation?
Mainly because the question says you're not meant to. But if you do solve it, and find values for a and b you'll end up with complex roots and it doesn't look pretty when you take the cube root of that. Essentially you're meant to have the final answer as -1 or something with a root 21 in it. I'll check and post the answers up when I get home.

4. ## Re: Find without solving?!

The equation is $x^2- 5x+ 8= 0$. If a and b are roots of that equation, then $(x- a)(x- b)= x^2- (a+ b)x+ ab= x^2- 5x+ 8$ so we must have a+ b= 5 and ab= 8. Now we can use $a+ b= (a^{1/3}+ b^{1/3})(a^{2/3}+ (ab)^{1/3}+ b^{2/3})$ which, with a+ b= 5, ab= 8 becomes $5= (a^{1/3}+ b^{1/3})(2+ a^{2/3}+ b^{2/3})$.

Now, look at the fact that [tex](a^{1/3}+ b^{1/3})^2= a^{2/3}+ 2(ab)^{1/3}+ b^{2/3}= 4+ a^{2/3}+ b^{2/3}. That is, $2+ a^{2/3}+ b^{2/3}= -2+ (a^{1/3}+ b^{1/3})^2$. Putting that into the previous equation, and letting $a^{1/3}+ b^{1/3}= X$ we have $5= X(-2+ X^2)$ or $X^3- 2X- 5= 0$.

5. ## Re: Find without solving?!

Thanks mate. I think I got it from there.

6. ## Re: Find without solving?!

We are still not to the bottom of this question.
Sorry HallsofIvy, but you lost a negative sign in your factorisation of $a+b,$ the middle term in the second bracket should be negative.
Following that through I think that the equation at the end should be $X^{3}-6X-5=0.$
This has an obvious solution of $X=-1,$ and the other two are then found to be $2.7913$ and $-1.7913$ (4dp).
I suppose that it is the positive root that we are looking for.