How would one simplify the cube root of $x^{64}$? Please show the solution step by step if possible. Been a while since I took math so please forgive the simplicity of this question... Thank you in advance.
Asked
Active
Viewed 119 times
0
-
One reason for this question would be to emphasise that the answer is not $x^4$ (which some might be trapped into thinking by $4^3=64$) – Mark Bennet Jan 21 '18 at 08:58
3 Answers
3
$\sqrt[3]{x^{64}}=\sqrt[3]{x^{21\cdot 3+1}}=x^{21}\sqrt[3]{x}$.
Alex Ravsky
- 90,434
-
1
-
-
@MohammadZuhairKhan Yes, it absolutely is. $\sqrt{x^2}=|x|$ but logic like this and above would have you believe that $\sqrt{x^2}=x^{\frac{2}{2}}=x$ – JMoravitz Jan 21 '18 at 04:58
-
-
1@JMoravitz The question is about a cubic root, not a square root, and $\sqrt[3]{x^3}=x$ independently on the sign of $x$. – Alex Ravsky Jan 21 '18 at 05:05
-
What is the result of the left-hand-side of your answer when $x=-1$ then? What is the result of the right-hand side when $x=-1$? What is the principal cubic root of $-1$? Is it real? Is the notation $\sqrt[3]{x}$ used for the realvalued root or the principal root? – JMoravitz Jan 21 '18 at 05:06
-
2@JMoravitz This is marked precalculus. Clearly we are not working with complex numbers. – Yakov Shklarov Jan 21 '18 at 05:09
-
Thank you. Can you explain how you got that answer? How does the 21 come out. Thanks – user522534 Jan 21 '18 at 05:09
-
@user522534 The answer follows from the general rules to deal with the powers. $21$ is the integer part of $64/3$. – Alex Ravsky Jan 21 '18 at 05:11
-
@AlexRavsky You should really elaborate some more on why $\sqrt[3]{x^{64}}=\sqrt[3]{x^{21\cdot 3+1}}=x^{21}\sqrt[3]{x}$ may be valid here, while the very similarly looking $\sqrt[6]{x^{68}}=\sqrt[6]{x^{11\cdot 6+2}} \ne x^{11}\sqrt[6]{x^2}$ is clearly not. – dxiv Jan 21 '18 at 05:18
-
So essentially we have cube root of x^7 * x^7 * x^7 * x^3 * x? Then we take out the 3 “x^7” ‘s adding up to 21 (since multiplying exponents is really adding them) and left with an “x” inside the cube root? Only thing I dont get is dont we have an x^3 left as well and do we take this out? Sorry, I know I am dumbing it down... – user522534 Jan 21 '18 at 05:19
-
@JMoravitz Even if we allow with complex roots, the value of the left hand side is the same as the value of the right hand side. – Alex Ravsky Jan 21 '18 at 05:19
-
What is $(-1)^{64}$? It is $1$. What is the principal cube root of $1$? It is again $1$. Meanwhile, what is $(-1)^{21}$? It is $(-1)$. What is the principal cube root of $(-1)$? It is $\frac{1}{2}+\frac{\sqrt{3}}{2}i$. What is their product? $-\frac{1}{2}-\frac{\sqrt{3}}{2}i$. Is $1$ equal to $-\frac{1}{2}-\frac{\sqrt{3}}{2}i$? No. Is your left hand expression equal to the right when $x=-1$? No. – JMoravitz Jan 21 '18 at 05:24
-
@user522534 No, we have a cube root of $x^{21}\cdot x^{21}\cdot x^{21}\cdot x$. We know, that $\sqrt[3]{y^3}=y$ and $\sqrt[3]{z\cdot t}=\sqrt[3]{z} \cdot \sqrt[3]{t}$ for any real $y,z,t$. So $\sqrt[3]{x^{21}\cdot x^{21}\cdot x^{21}\cdot x}= \sqrt[3]{x^{21}\cdot x^{21}\cdot x^{21}}\cdot \sqrt[3]{x}= x^{21}\cdot \sqrt[3]{x}$. – Alex Ravsky Jan 21 '18 at 05:26
-
@JMoravitz What do you mean by the principal cube root? And why the principal cube root of $1$ is $1$, whereas the principal cube root of $(-1)$ is $\frac 12+\frac{\sqrt 3}2i$? – Alex Ravsky Jan 21 '18 at 05:29
-
THANK YOU SO MUCH. My brain just connected the dots. Total brain malfunction... Obviously math is not my forte! Greatly appreciated! – user522534 Jan 21 '18 at 05:32
-
@JMoravitz I have to go to my university, so answering your possible future question I tell that for real $x$ and complex roots at both sides we have sets ${1,\varepsilon,\varepsilon^2}|x|^{64/3}$, where $\varepsilon\ne 1$ is the complex cubic root of $1$. – Alex Ravsky Jan 21 '18 at 05:42
-
1@AlexRavsky the principal $n$'th root of a number is the $n$'th root whose argument is closest to zero. Writing a number in polar form $re^{i\theta}$ where $r$ is non-negative real and $0\leq\theta<2\pi$, one has $\sqrt[n]{re^{i\theta}}=\sqrt[n]{r}\cdot e^{i\theta/n}$ where $\sqrt[n]{r}$ is defined as usual for non-negative real numbers. See $n$'th root on wikipedia and $n$'th complex roots – JMoravitz Jan 21 '18 at 05:43
-
OK, I should go now, so I look at your links later. Since I work rather in algebra than in complex analysis, for me complex roots constitute a set and I don’t demand that the expression $\sqrt[3]{x}$ is reserved to represent some specific element of this set. Thank you for you attentivity, I didn’t consider so deep issues when I wrote my answer. – Alex Ravsky Jan 21 '18 at 05:51
3
$x^{64}=x^{\frac{64}{3}\times3}$.
Hence, $^3\sqrt{x^{64}}=x^{\frac{64}{3}}$.
Even when $x$ is negative, it yields the same results at there is an even power of $64$ which cancels out the negative.
On the other hand, the similar looking $x^\frac{5}{3}$ does not hold for negative values as $5$ is a odd power, and hence the negative sign is not cancelled.
Hence, we can form a generalisation for all negative $x$: for $x^\frac{a}{3}$, if $a$ is even, the result is positive. If $a$ is odd, the result is negative.
QuIcKmAtHs
- 1,451
-
-
-
mmm... no. If you want to be pedantic, when you use non-integer rational exponents you actually get a multivalued expression, or if you want to make it single-valued you use the denominator of the rational number first rather than the numerator, specifically using the principal root. Your expression $x^{\frac{64}{3}}$ under such a convention yields a nonreal complex number for $x=-1$ – JMoravitz Jan 21 '18 at 05:14
-
-
@XcoderX You should really elaborate some more on why $^3\sqrt{x^{64}}=x^{\frac{64}{3}}$ may be valid here, while the very similarly looking $\sqrt[6]{x^{10}}=x^{\frac{5}{3}}$ is clearly not. – dxiv Jan 21 '18 at 05:22
-
-
@XcoderX "I tried $x=-1$ and got $1$." Wolfram-alpha disagrees as do I and pen-and-paper calculations should verify. Notice below the entry box a phrase "Assuming the principal root | Use the real-valued root instead" or the opposite "Assuming the real-valued root | Use the principal root instead." Remembering that the convention is and should be to use the principal root for this notation, one shouldn't get a real valued answer to $(-1)^{\frac{64}{3}}$. – JMoravitz Jan 21 '18 at 05:55
-
As per your edit in regards to the $\sqrt[6]{x^{10}}$ vs $x^{\frac{5}{3}}$ side question that dxiv tried to pose to get you to think about the underlying issues at work here, you seemed to miss the point. Even if you were to use the real-valued root instead of the principal root, according to your logic you might say $\sqrt[6]{x^{10}}=x^{\frac{10}{6}}=x^{\frac{5}{3}}$, but by plugging in $-1$ you have a positive number on the far left but a negative number on the far right, or $\sqrt{x^2}=x^{\frac{2}{2}}=x$, again with $x=-1$ gives a positive number on the left but a negative number on right. – JMoravitz Jan 21 '18 at 06:09
-
The question to think about here is: Is $\sqrt[b]{x^a}=x^{\frac{a}{b}}$ actually a valid simplification for all $x$ including negative values, and given $\frac{a}{b}=\frac{c}{d}$, is $x^{\frac{a}{b}}=x^{\frac{c}{d}}$ actually a valid simplification for all $x$ including negative values? The above examples should tell you that at least one of these two simplifications is incorrect. Which, and why? – JMoravitz Jan 21 '18 at 06:13
-1
You need to be careful. Assuming that $x$ is real, one has $x^{64}$ is always a non-negative real number, and the cube root of a non-negative real number will again be a non-negative real.
$\sqrt[3]{x^{64}}=|x|^{\frac{64}{3}}=|x^{21}\cdot \sqrt[3]{x}|$
Without such care, one can arrive at incorrect results such as $\sqrt{x^2}=x$, which would imply things such as $1=-1$ (by plugging in $-1$ in for $x$)
JMoravitz
- 79,518
-
1As Alex Ravsky has pointed out, for odd powers this is unnecessary. The equation $\sqrt[3]{a^3} = a$ holds for all real $a$. – Yakov Shklarov Jan 21 '18 at 05:07
-
I am not taking issue with $\sqrt[3]{a^3}=a$. I am taking issue with all of the other erroneous simplifications being made @YakovShklarov. – JMoravitz Jan 21 '18 at 05:10
-
@E.H.E. $\sqrt{x^2}=|x|$ and not $\sqrt{x^2}=x$ is a classic example of why care needs to be taken in such things which one can see in problems such as Why $\sqrt{-1\times -1}\neq \sqrt{-1}^2?$. And I take back what I said earlier. I do take issue with saying $\sqrt[3]{a^3}=a$. That would be true if we were using the realvalued cube root rather than the more standard convention of using the principal cube root where such a simplification is false. $\sqrt[3]{(-1)^3}=\sqrt[3]{-1}=\frac{1}{2}+\frac{\sqrt{3}}{2}i$ – JMoravitz Jan 21 '18 at 05:27
-
1@E.H.E Both other answers handwave that $,(x^a)^b=x^{a \cdot b},$, which is not true in general. Even if that works in this case, any answer that doesn't emphasize why it happens to work here, and where else it does not work, is a dangerously incomplete answer. Whoever downvoted this answer did not understand that point, and is at risk of falling for any one of those $,-1=(-1)^{2/2}=\sqrt{(-1)^2}=1,$ perennial questions that get posted on MSE every other week. – dxiv Jan 21 '18 at 05:40