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I need to make the fraction inside of the roots become the same with one and another so I can easily sum it in the end of the process for another calculation. $$v_{1}=\sqrt[4]{24}=2\sqrt[4]{\frac 32}$$ $$v_{2}=\sqrt[4]{\frac 32}$$ $$v_{3}=\sqrt[4]{\frac {8}{27}}=\frac 23\sqrt[4]{\frac 32}$$ $$v_{4}=\sqrt[4]{\frac {3}{32}}=\frac 12\sqrt[4]{\frac 32}$$ Since the fractions were simple, I could easily calculate and make it all became same fraction in each root. But I can't solve the problem below because the numbers are bigger. $$v_{1}=\sqrt[4]{\frac {1}{245}}$$ $$v_{2}=\sqrt[4]{\frac 76}$$ $$v_{3}=\sqrt[4]{\frac {14}{9}}$$ $$v_{4}=\sqrt[4]{135}$$ Could anyone kindly help me to solve this problem or let me know the "official" method/formula to simplify it?

Thank you

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Are you adding the roots or are you adding numbers within the roots? If you are adding fractions within the roots you add them in the normal way. There is not an easy way of adding roots. For example, 3*(17)^(1/2) + 26^(1/2) does not have an easy way of solving it. But, if you have a coefficient times the nth root of x plus some other coefficient times the nth root of x, you add the coefficients. If you have six of something, and you add to it 10 of something, you have sixteen of something. The somethings don't matter. They can be roots, powers, apples, iPhone 6s's, etc.

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  • $\begingroup$ The reason I want to simplify it is because after all that, I need to sum all of the $v$ and later, I need to calculate one by one $\frac {v_{1}}{V}$, and so on.. So I just thought it would be easier to calculate if the fraction inside of each roots are the same. Anyway, thank you for the answers! $\endgroup$ – Putri Alifiar Apr 6 '17 at 14:47
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I don't think it's possible. In your first example, the exponent on $3$ inside the radical is congruent to $1$ modulo $4$ and the exponent on $2$ is congruent to $3$ modulo 4 in every instance.

But in your second example, in $v_2$, $3$ appears to the $-1$ power, but in $v_3$ it appears to the $-2$ power. There are similar problems with all the other primes. Note that $4$ is the modulus that matters, because you're taking $4$th roots.

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    $\begingroup$ Equivalently, just note that (for example) the quotient $\sqrt[4]{14/9} / \sqrt[4]{7/6} = \sqrt[4]{4/3}$, which is irrational. $\endgroup$ – Connor Harris Apr 6 '17 at 14:38
  • $\begingroup$ The reason I want to simplify it is because after all that, I need to sum all of the $v$ and later, I need to calculate one by one $\frac {v_{1}}{V}$, and so on.. So I just thought it would be easier to calculate if the fraction inside of each roots are the same. Anyway, thank you for the answers! $\endgroup$ – Putri Alifiar Apr 6 '17 at 14:47
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In your first example instead of keeping $\frac 32$ let's just eliminate the denominators by factorizing them out of the root operator.

$\begin{cases} v_1=\sqrt[4]{24} \\ v_2=\sqrt[4]\frac 32=\frac 12\sqrt[4]{3.2^3}=\frac 12\sqrt[4]{24}\\ v_3=\sqrt[4]\frac 8{27}=\frac 13\sqrt[4]{8.3}=\frac 13\sqrt[4]{24}\\ v_4=\sqrt[4]\frac 3{32}=\frac 12\sqrt[4]{\frac 32}=\frac 14\sqrt[4]{3.2^3}=\frac 14\sqrt[4]{24}\\ \end{cases}$

And we see that it works because when rationalizing all this, then this is the same $\sqrt[4]{24}$ that appears.

Now if we do the same with your numbers, we have no such luck

$\begin{cases} v_1=\sqrt[4]{\frac 1{245}}=\sqrt[4]{\frac 1{5.7^2}}=\frac 1{35}\sqrt[4]{5^3.7^2}=\frac 1{35}\sqrt[4]{6125} \\ v_2=\sqrt[4]\frac 76=\frac 16\sqrt[4]{7.2^3.3^3}=\frac 16\sqrt[4]{1512}\\ v_3=\sqrt[4]\frac {14}9=\frac 13\sqrt[4]{14.3^2}=\frac 13\sqrt[4]{126}\\ v_4=\sqrt[4]{135}\\ \end{cases}$

And these are not much more reductible since there are no power of $4$ inside the root operator.

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