# Finding Rational numbers

Find rational numbers a and b such that:

$$\left(7 + 5\sqrt2\right)^{\frac13} = a + b \sqrt2$$

Thank you

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you state a and b are rational in your question, but I assume they are irrational by the title? – David Sep 12 '13 at 8:27
No comprende plain text. Habla mathjax? :) – David H Sep 12 '13 at 8:29
edited it thanks david – Charlie Sep 12 '13 at 8:30
Well, take both sides to the power of 3 calculate the right hand side (i.e. $(a+b\sqrt{2})^3$). Comparing the coefficients (of $1$ and $\sqrt{2}$), you get two equations where you can immediately see a very simple choice for a and b. – roman Sep 12 '13 at 8:43
When you consider that there isn't one pair (a,b), you could follow through with setting a or b to something, and solving for the other. This will give you a pair of rational numbers, but it won't give you the set of all rational numbers. – David Sep 12 '13 at 8:47

Hint: Cubing both sides you have $$7+5\sqrt{2}=(a^3+6ab^2)+(2b^3+3a^2b)\sqrt{2}$$ Since $a,b$ are rational, equating the rational and irrational parts of both sides, you get $$a^3+6ab^2=7\\ 2b^3+3a^2b=5$$ Now, from the first equation you get $$b^2=\frac{7-a^3}{6a}$$. Put that in the second equation to get $$b^2(2b^2+3a^2)^2=25\\ \Rightarrow (7-a^3)\left(\frac{7-a^3}{3a}+3a^2\right)^2=150a\\ \Rightarrow (7-a^3)\left(8a^3+7\right)^2=9\cdot150\cdot a^3$$Now, let $a^3=x$ Then you need to solve the cubic equation $$(7-x)(8x+7)^2=1350x\\ \Rightarrow (7-x)(64x^2+112x+49)=1350x\\ \Rightarrow 64x^3-336x^2+615x-343=0$$ Amazingly, Wolfram alpha gives one real solution to this $x=1$! which corresponds to the real solution of $a=1$ and hence $b=1$ which are obviously rational and hence you are done.

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It's $2b^3+3a^2b=5$, not 6. – roman Sep 12 '13 at 9:15
Yes, yes, I found that typo and subsequent disaster as soon as I posted it. I have edited it now :-). – Samrat Mukhopadhyay Sep 12 '13 at 9:18
thank you very much guys – Charlie Sep 12 '13 at 12:19

Note that

$$\sqrt2 b+a = ({7+5 \sqrt2)})^{\frac{1}{3}}\\ \implies (\sqrt2 b+a)^3 = 7+5 \sqrt{2}\\ \implies a^3 + 3b\sqrt{2}a^2 + 6b^2a + 2b^3\sqrt{2} = 7+5 \sqrt{2}\\ \implies (a^3+6b^2a) + (3ba^2+2b^3)\sqrt{2} = 7+5 \sqrt{2}\\$$

This yields two equations:

$$a^3+6b^2a = 7\\ 3ba^2+2b^3 = 5$$

There is an immediate, obvious choice for $a$ and $b$ here.

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check this:

$\sqrt2 b+a = ({7+5 \sqrt2)})^{\frac{1}{3}}$

Isolate terms with b to the left hand side. Subtract a from both sides:

$\sqrt2 b = ({7+5 \sqrt2)})^{\frac{1}{3}}-a$

Solve for b. Divide both sides by sqrt(2):

$b = \frac{({7+5 \sqrt2)})^{\frac{1}{3}}- a}{\sqrt2}$

Can you take it from here?

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$a,b$ have to be rationals right? – Samrat Mukhopadhyay Sep 12 '13 at 8:50
I couldn't figure that one out from the comments vs. title...let the asker clarify – DanielY Sep 12 '13 at 8:55
well i kind of did that expanded then isolated terms and found one term in terms of another i dont know if what am i doing is right – Charlie Sep 12 '13 at 8:55
from this point, you need to find $a,b$ that must be rational. Just pick up a rational number a and solve the equation to find b. If b is also rational, you're done – DanielY Sep 12 '13 at 8:58
when i cube both sides i get this: – Charlie Sep 12 '13 at 9:01