Is there faster way to calculate $\frac{4^{0.75}}{1+\sqrt2+\sqrt3}+9^{0.25}$?

We have $$\dfrac{4^{0.75}}{1+\sqrt2+\sqrt3}+9^{0.25}=\dfrac{2\sqrt2}{1+\sqrt2+\sqrt3}+\sqrt3$$ To evaluate this, the standard approach is multiplying the fraction by $$\dfrac{1-\sqrt2-\sqrt3}{1-\sqrt2-\sqrt3}$$ and then we left with one square root in the denominator and by multiplying the fraction by the conjugate of the denominator again we get rid of all square roots in denominator. but my question is : Is there another approach (preferably faster one) to evaluate the expression?

I also tried using the common denominator and get this:

$$\dfrac{2\sqrt2+\sqrt3+\sqrt6+3}{1+\sqrt2+\sqrt3}$$ It seems I can't proceed from here without using the previous method.

• No................... Mar 30 '21 at 18:20
• It's not clear what kind of answer you are looking for. Mar 30 '21 at 18:35
• use FullSimplify[2 Sqrt/(1 + Sqrt + Sqrt) + Sqrt] in Mathematica to get 1+Sqrt. Mar 30 '21 at 18:44
• @Somos How is that possible. There is no $\sqrt{3}$ at the end?
– Aqua
Mar 30 '21 at 18:48

Start with $$\frac{2\sqrt2}{1+\sqrt2+\sqrt3}+\sqrt3. \tag{1 }$$ Put them both in common denominator $$\frac{2\sqrt2}{1+\sqrt2+\sqrt3} + \frac{\sqrt3(1+\sqrt2+\sqrt3)}{1+\sqrt2+\sqrt3}. \tag{2}$$ Add the two fractions $$\frac{2\sqrt2 + (\sqrt3+\sqrt6+3)}{1+\sqrt2+\sqrt3}.\tag{3}$$ Collect the numerator $$\frac{3 + 2\sqrt2 + \sqrt3+\sqrt6}{1+\sqrt2+\sqrt3}.\tag{4}$$ Take out $$1$$ from the fraction $$1 + \frac{\sqrt2 + 2 + \sqrt6}{1+\sqrt2+\sqrt3}. \tag{5}$$ Now recognize that the numerator is divisible by the denominator. $$1 + \sqrt2. \tag{6}$$

In general, using the same reasoning we get $$\frac{(2\!+\!a\!-\!n)\!+\!(a\!+\!1)\sqrt2 \!+\!(a\!-\!1)\sqrt{n}} {1+\sqrt2+\sqrt{n}}\!+\!\sqrt{n} \!=\! a \!+\! \sqrt2.$$ Make the substitutions $$\,a=1,n=3\,$$ to get our special case.

Since $$4^{0.75}=4^\frac34=\left(2^2\right)^\frac34=2^\frac32=\sqrt{2^3}=2\sqrt2$$

and

$$9^{0.25}=9^\frac14=\left(3^2\right)^\frac14=3^\frac12=\sqrt3\;,$$

it follows that

$$\dfrac{4^{0.75}}{1+\sqrt2+\sqrt3}+9^{0.25}=\dfrac{2\sqrt2}{1+\sqrt2+\sqrt3}+\sqrt3\;.$$

Moreover,

$$\dfrac{2\sqrt2}{1+\sqrt2+\sqrt3}+\sqrt3=\dfrac{\left(3+2\sqrt2\right)-3}{1+\sqrt2+\sqrt3}+\sqrt3=$$

$$=\dfrac{\left(1+\sqrt2\right)^2-\left(\sqrt3\right)^2}{1+\sqrt2+\sqrt3}+\sqrt3=$$

$$=\dfrac{\left(1+\sqrt2+\sqrt3\right)\left(1+\sqrt2-\sqrt3\right)}{1+\sqrt2+\sqrt3}+\sqrt3=$$

$$=\left(1+\sqrt2-\sqrt3\right)+\sqrt3=1+\sqrt2\;.$$

• You are just repeating what the OP already said. Mar 30 '21 at 18:38
• Thanks, But I already knew this. my question was how to calculate $\dfrac{2\sqrt2}{1+\sqrt2+\sqrt3}+\sqrt3$ without using conjugates method. Mar 30 '21 at 18:38
• I have calculated your expression without using conjugates method. Mar 30 '21 at 19:12
• Why have you downvoted my answer? Is there any mistakes? If yes, what are my mistakes? Mar 30 '21 at 19:22
• I explained why I downvoted your answer. Since then, you have edited it. Mar 30 '21 at 19:25