Cube root and short multiplication formulas

Question

I've got this math assignment. I'm stuck with it and I'd love it if someone could help me out with it, thanks! :)

$$\sqrt[3]{5\sqrt2 + 7} - \sqrt[3]{5\sqrt2-7}=2$$

Edit: So far I've tried using short multiplication formules, simplifying the roots or doing the opposite to suite the $a^2 + 2ab + b^2$, but so far I think I was aproaching this the wrong way. My bad for not following some rules, my first post here sorry. Edit 2: Got it all, explained it to the class. Huge thanks to everyone this community is really awesome! :)

Answer 1

Hint:  let $a=\sqrt[3]{5\sqrt2 + 7}, b =\sqrt[3]{5\sqrt2-7}$ then $a^3-b^3=14$ and $ab=1\,$, so:

$$(a-b)^3=a^3-b^3-3ab(a-b)=14-3(a-b)$$

Now look for the real solution(s) of the equation $x^3+3x-14=0\,$.

Answer 2

Hint:

$$5\sqrt 2+7=(\sqrt2+1)^3.$$

Answer 3

Here's how you can do it - take equation to be LHS $= x = a-b$. Then use identity $(a-b)^3 = a^3 - b^3 -3ab(a-b)$ where $a = (5\sqrt 2 +7)^{1/3}$ and $b= (5\sqrt 2 - 7)^{1/3} $. Then solving, $ a^3 - b^3 = (5\sqrt 2 +7) - (5\sqrt 2 - 7) = 14 , ab = (50 - 49)^{1/3} = 1^{1/3}=1 $ (apply $ (x+y)(x-y) = x^2 - y^2 )$ :- you will get $ x^{3}=14 - 3x $ whose only real solution is 2.

Answer 4

Ask what $5\sqrt2 + 7$ and $5\sqrt2-7$ would look like as cubes of something. Envision that:

$$5\sqrt2 + 7=\left(a+b \sqrt{2}\right)^3$$

This implies that $$7+5\sqrt2=\left(a^3+6ab^2\right) + \left(3a^2b+2b^3\right)\sqrt{2}$$ So it must be that $$\begin{align} \begin{cases} 7&=a^3+6ab^2 &=a(a^2+6b^2) \\ 5&= 3a^2b+2b^3 &= b(3a^2+2b^2)\end{cases}\end{align}$$

a quick inspection, using the convenient fact that $7$ and $5$ are primes, leads us to $$[a=b=1] \implies $$ $$7+5\sqrt2 =\left(1+ \sqrt{2}\right)^3$$

Now try it for

$$5\sqrt2 - 7=\left(c+d \sqrt{2}\right)^3$$

You can bet that

$$[c=-1 \ \text{and} \ d=1] \implies$$

$$-7+5\sqrt2 =\left(-1+\sqrt{2}\right)^3$$