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[ASK] Exponents and Roots Simplification problem

Monoxdifly

Well-known member
Aug 6, 2015
267
The result of \(\displaystyle \frac{7x-\frac92\sqrt[6]{y^5}}{\left(x^{\frac56}-6y^{-\frac13}\right)x^{-2}}\) for x = 4 and y = 27 is ....
a. \(\displaystyle \left(1+2\sqrt2\right)9\sqrt2\)
b. \(\displaystyle \left(1+2\sqrt2\right)9\sqrt3\)
c. \(\displaystyle \left(1+2\sqrt2\right)18\sqrt3\)
d. \(\displaystyle \left(1+2\sqrt2\right)27\sqrt2\)
e. \(\displaystyle \left(1+2\sqrt2\right)27\sqrt3\)

I got stuck at \(\displaystyle \frac{56-3^2\left(3^{\frac52}\right)}{2^-{\frac43}-2^{-2}}\) and don't know how to continue to reach one of the options. By the way, don't you think that 7 is suspicious?
 

skeeter

Well-known member
MHB Math Helper
Mar 1, 2012
613
North Texas
I simplified the original expression to $\dfrac{4(56-81\sqrt{3})}{\sqrt[3]{4}-1}$

Note evaluation of the original numerator ...

$28 - \frac{9}{2} \cdot 27^{5/6} \approx -42.1481$

evaluation of the denominator ...

$(4^{5/6} - 6\cdot 27^{-1/3}) \cdot \frac{1}{16} \approx 0.0734$

Division would yield a negative value. Clearly, all choices are positive. Maybe a typo or other error with the original expression?
 

Monoxdifly

Well-known member
Aug 6, 2015
267
Division would yield a negative value. Clearly, all choices are positive. Maybe a typo or other error with the original expression?
According to you, what part of the question should be gotten rid of to reach one of the options?
 

skeeter

Well-known member
MHB Math Helper
Mar 1, 2012
613
North Texas
According to you, what part of the question should be gotten rid of to reach one of the options?
Something is in error with the original expression. Could be an omission, a sign error, a coefficient error, an error in one or more exponents, or a combination of the aforementioned.

You can make an attempt to reverse engineer the original expression to fit one of the choices if you have the time and the inclination.
 

Monoxdifly

Well-known member
Aug 6, 2015
267
You can make an attempt to reverse engineer the original expression to fit one of the choices if you have the time and the inclination.
I don't. Especially when I don't even know what's the supposed right answer. Not gonna waste my time reverse-engineering up to five options. Gonna raise my white flag.
 

DavidCampen

Member
Apr 4, 2014
64
Southern California
I simplified the original expression to $\dfrac{4(56-81\sqrt{3})}{\sqrt[3]{4}-1}$
I can duplicate Skeeters answer. I wouldn't have had the patience to try simplifying the original expression if Skeeter hadn't already given the answer. It was very tedious and I had to go over it several times to correct mistakes. They must make software to do this.
 

Wilmer

In Memoriam
Mar 19, 2012
376
Ottawa Ontario Canada
Mr.Fly, you posted a "bad" expression, obviously.

Go stand in the corner for y minutes...
 

Wilmer

In Memoriam
Mar 19, 2012
376
Ottawa Ontario Canada
By the way, don't you think that 7 is suspicious?
Well, if that 7 was 20, then you'd be "in the range", with result = ~134.17,
between 3rd choice of ~119.36 and 4th choice of ~146.18.
 

Wilmer

In Memoriam
Mar 19, 2012
376
Ottawa Ontario Canada
Holy crappy!

So numerator: 7 * x * 9^(1/2) * y^(5/6)
denominator : [x^(5/4) - 6 * y^(-1/3)] * x^(1/2)

which matches the 5th choice...

Remind me not to look at any of your future equations :)
 

DavidCampen

Member
Apr 4, 2014
64
Southern California

Monoxdifly

Well-known member
Aug 6, 2015
267