how to prove the value of f = ( a + ((a+1)/3)*sqrt(((8*a-1)/3)^(1/3)) + nthroot(( a - ((a+1)/3)*sqrt(((8*a-1)/3), 3) is always 1

[sallyyu2022]

how to prove the value of f = ( a + ((a+1)/3)*sqrt(((8*a-1)/3)^(1/3)) + nthroot(( a - ((a+1)/3*sqrt(((8*a-1)/3),3) is always 1

I try to use X=a+b, Y=a-b ,when b= ((a+1)/3)*sqrt(((8*a-1)/3)^(1/3) to simplify the function and then prove that (X+Y)^3=1. However, I got stuck when I come to the expansion equation.

I assume that binomial theorem could be helpful to some steps in the process, but I am not sure.

 

[stapel]

how to prove the value of f = ( a + ((a+1)/3)*sqrt(((8*a-1)/3)^(1/3)) + nthroot(( a - ((a+1)/3*sqrt(((8*a-1)/3),3) is always 1

What value are you assigning to [imath]n[/imath]? What is the meaning of the comma before the last [imath]3[/imath]?

If possible, could you please post a picture of the expression in question?

I try to use X=a+b, Y=a-b ,when b= ((a+1)/3)*sqrt(((8*a-1)/3)^(1/3) to simplify the function and then prove that (X+Y)^3=1. However, I got stuck when I come to the expansion equation.

Could you post a picture of at least one of your attempts, so that we can see what you're doing?

I assume that binomial theorem could be helpful to some steps in the process, but I am not sure.

Why do you assume that the binomial theorem would be helpful?

Thank you!

 

[sallyyu2022]

This screenshot below show the original fucntion

I tried to observe the different results in matlab when a could be any different real number while larger or equal to 1/8 , so I offered a script in matlab. Maybe it cause some confusions.
I try to prove the cubic value of f is equal to 1 so that the funtion could be simplfied.
Firstly, I set X=a+b, Y=a-b ,and b=.

 

[sallyyu2022]

Then the expansion equation of the cubic value of f is , where is equal to 2a. However, I got stuck by how to deal with

 

[Dr.Peterson]

Then the expansion equation of the cubic value of f is View attachment 36942, where View attachment 36943is equal to 2a. However, I got stuck by how to deal with View attachment 36944

Those terms, together, will be [imath]3XY(X+Y)[/imath]. Do the multiplication and the addition, and things will work out. If you're still stuck there, show us that part of your work.

By the way, I don't think it's standard to say "the cubic value of f"; you mean the cube of f, or the value of f^3. (And that is a good strategy.)

 

[Dr.Peterson]

This screenshot below show the original fucntion
View attachment 36938
Firstly, I set X=a+b, Y=a-b ,and b=View attachment 36939.

I realize on second glance that you didn't type what you meant; what I assume you meant is

Let X^3 = a + b, and Y^3 = a - b.​

Then f^3 = (X + Y)^3 = X^3 + 3X^2Y + 3XY^2 + Y^3 = (X^3 + Y^3) + 3XY(X + Y). Since we don't have such simple expressions for X and Y themselves (they're cube roots), the last part isn't as nice as it initially seemed. There's more work to do.

 

[blamocur]

Try replacing [imath]s = \sqrt{\frac{8a-1}{3}}[/imath]

 

[limiTS]

Let [imath]\displaystyle x=\sqrt[3]{a+\frac{a+1}{3}\sqrt{\frac{8a-1}{3}}}[/imath] and [imath]\displaystyle y=\sqrt[3]{a-\frac{a+1}{3}\sqrt{\frac{8a-1}{3}}}[/imath]

Then, [imath]\displaystyle x^3=a+\frac{a+1}{3}\sqrt{\frac{8a-1}{3}}[/imath] and [imath]\displaystyle y^3=a-\frac{a+1}{3}\sqrt{\frac{8a-1}{3}}[/imath]

Hint: the sum of cubes and difference of cubes formulae are:
[imath]x^3+y^3=\left(x+y\right)\left(x^2-xy+y^2\right)[/imath]
[imath]x^3-y^3=\left(x-y\right)\left(x^2+xy+y^2\right)[/imath]

Spoiler

[math]\begin{aligned} x^3+y^3&=\left(x+y\color{black}\right)\left(x^2-xy+y^2\color{black}\right)\\&=\left(x+y\color{black}\right)\left(x^2\color{red}{+2xy}\color{black}+y^2-xy\color{red}{-2xy}\color{black}\right)\\ &=\left(x+y\color{black}\right)\left[\left(x+y\right)^2-3xy\right]\\ &=\left(x+y\color{black}\right)^3-3xy(x+y)\\ 0&=\underbrace{\left(x+y\color{black}\right)^3-\color{blue}x^3}_{\text{difference of cubes}}\color{black}-3xy(x+y)-\color{blue}y^3\\ &=(x+y-\color{blue}{x}\color{black})\left[(x+y)^2+(x+y)\color{blue}x\color{black}+\color{blue}{x}\color{black}^2\right]-3xy(x+y)-\color{blue}y^3\\ &=y\left[(x+y)^2+x(x+y)+x^2\right]+y\left[-3x(x+y)-y^2\right]\\ &=y\left[(x+y)^2+x(x+y)+x^2-3x(x+y)-y^2\right]\\ &=y\left[(x+y)^2-2x(x+y)+x^2-y^2\right]\\ &=y\left[(x+y)^2-2x(x+y)+(x+y)(x-y)\right]\\ &=y\left[(x+y)^2-\left\{2x(x+y)-(x+y)(x-y)\right\}\right]\\ &=y\left[(x+y)^2-(x+y)\left\{2x-(x-y)\right\}\right]\\ 0&=y\underbrace{\left[(x+y)^2-(x+y)\left\{x+y\right\}\right]}_{=0}\\ \end{aligned}[/math]
Now we know [imath]\displaystyle y^3=0=a-\frac{a+1}{3}\sqrt{\frac{8a-1}{3}}[/imath]
Solve: [imath]a=\dfrac{1}{2}[/imath]

Plug [imath]a=\dfrac{1}{2}[/imath] into [imath]\displaystyle x=\sqrt[3]{a+\frac{a+1}{3}\sqrt{\frac{8a-1}{3}}}=\sqrt[3]{1}[/imath], so [imath]x=1[/imath]

[imath]x+y=1+0=1[/imath], always.

 

[Dr.Peterson]

Let [imath]\displaystyle x=\sqrt[3]{a+\frac{a+1}{3}\sqrt{\frac{8a-1}{3}}}[/imath] and [imath]\displaystyle y=\sqrt[3]{a-\frac{a+1}{3}\sqrt{\frac{8a-1}{3}}}[/imath]

Then, [imath]\displaystyle x^3=a+\frac{a+1}{3}\sqrt{\frac{8a-1}{3}}[/imath] and [imath]\displaystyle y^3=a-\frac{a+1}{3}\sqrt{\frac{8a-1}{3}}[/imath]

Hint: the sum of cubes and difference of cubes formulae are:
[imath]x^3+y^3=\left(x+y\right)\left(x^2-xy+y^2\right)[/imath]
[imath]x^3-y^3=\left(x-y\right)\left(x^2+xy+y^2\right)[/imath]

The step at which you solve for y, and then for a, is wrong. The expression is in fact equal to 1 for any (sufficiently large) value of a, not just for one value.

 

[limiTS]

The step at which you solve for y, and then for a, is wrong. The expression is in fact equal to 1 for any (sufficiently large) value of a, not just for one value.

Realized the error (on my own). I was going to edit my post, but time expired. I should have said that y can equal anything, which includes zero. So y=0 gets only one solution for a, and it's not a proof.

 

[blamocur]

The step at which you solve for y, and then for a, is wrong. The expression is in fact equal to 1 for any (sufficiently large) value of a, not just for one value.

Actually for [imath]1/8 \leq a \leq 1/2[/imath] if we stick to real numbers.

 

[Dr.Peterson]

Actually for [imath]1/8 \leq a \leq 1/2[/imath] if we stick to real numbers.

No, it's still defined (and real) past 1/2. Are you reading the cube roots as square roots?

I was intentionally not precise, but was observing that its domain starts at 1/8; one of the first things I'd done was to graph the function and see that it really is constant on that domain:

​

 

[blamocur]

Actually for [imath]1/8 \leq a \leq 1/2[/imath] if we stick to real numbers.

Ooops, I forgot that one can compute cube roots of negative values My Python script refused to do so, and I wasn't thinking clearly.

 

[limiTS]

OK, I think I got it this time.

As before, [imath]\displaystyle x=\sqrt[3]{a+\frac{a+1}{3}\sqrt{\frac{8a-1}{3}}}[/imath] and [imath]\displaystyle y=\sqrt[3]{a-\frac{a+1}{3}\sqrt{\frac{8a-1}{3}}}[/imath]

Then, [imath]\displaystyle x^3=a+\frac{a+1}{3}\sqrt{\frac{8a-1}{3}}[/imath] and [imath]\displaystyle y^3=a-\frac{a+1}{3}\sqrt{\frac{8a-1}{3}}[/imath]. We ultimately want to solve for [imath]x+y[/imath]

Note that [imath]x^3+y^3=2a[/imath] and [imath]\displaystyle xy=\sqrt[3]{a^2-\left(\frac{a+1}{3}\sqrt{\frac{8a-1}{3}}\right)^2}=\sqrt[3]{\frac{-8a^3+12a^2-6a+1}{27}}=-\frac{2a-1}{3}[/imath]

Spoiler

The sum of cubes can be expanded:
[imath]\begin{aligned}x^3+y^3&=2a\\\left(x+y\right)\left(x^2-xy+y^2\right)&=\\\left(x+y\right)\left(x^2+2xy+y^2-xy-2xy\right)&=\\\left(x+y\right)\left[\left(x+y\right)^2-3\underbrace{xy}_{\text{sub. in }xy}\right]&=\\\left(x+y\right)\left[\left(x+y\right)^2-3\left(-\frac{2a-1}{3}\right)\right]&=\\\left(x+y\right)\left[\left(x+y\right)^2+\left(2a-1\right)\right]&=\\\left(x+y\right)^3+\left(2a-1\right)\left(x+y\right)&=2a\\\left(x+y\right)^3+\left(2a-1\right)\left(x+y\right)-2a&=0\\\underbrace{\left(x+y\right)^3-1}_{\text{difference of cubes}}+\left(2a-1\right)\left(x+y\right)-2a+1&=\\\left(x+y-1\right)\left[(x+y)^2+(x+y)+1\right]+\left(2a-1\right)\left(x+y\right)-(2a-1)&=\\\left(x+y-1\right)\left[(x+y)^2+(x+y)+1\right]+\left(2a-1\right)\left[\left(x+y\right)-1\right]&=\\\left(x+y-1\right)\left[(x+y)^2+(x+y)+1+(2a-1)\right]&=\\\left(x+y-1\right)\left[(x+y)^2+(x+y)+2a\right]&=0\\\end{aligned}[/imath]

We can say that [imath](x+y-1)=0[/imath] and thus, [imath]\color{red}x+y=1[/imath], as the question wanted us to prove.
OR
[imath]\displaystyle (x+y)^2+(x+y)+2a=0\rightarrow s^2+s+2a=0\rightarrow s=x+y=\frac{-1\pm\sqrt{1-4(2a)}}{2}[/imath]

Observing that [imath]\displaystyle\sqrt{\frac{8a-1}{3}}\geq0[/imath], to stay in the reals, the smallest value for a is [imath]\dfrac{1}{8}[/imath]. For any [imath]a>\dfrac{1}{8}[/imath], the discriminant above is negative and we're no longer in the reals.