cubedroot250+cubedroot54-cubedroot16 ( i hope i wrote that right? @SolomonZelman lol
@MaxwellSmart
Yes, you did:) \(\normalsize\color{blue}{ \sqrt[3]{250}+\sqrt[3]{54}-\sqrt[3]{16} }\)
yes thats it!
Bam!!!!!!
\(\LARGE \sqrt[3]{250} + \sqrt[3]{54} - \sqrt[3]{16} \)
\(\normalsize\color{blue}{ \sqrt[3]{5^3 \times 2}+\sqrt[3]{3^3 \times 2 }-\sqrt[3]{2^3 \times 2} }\)
Almost a sequence :)
sorry guys i dont write it out like that...i dont have the sqrt symbol :(
In order to combine the roots together, they have to be roots of the same number.
Equation editor has it:)
1) Click ALT and Hold it, 2) Click 2 5 1 on the number pad, on the bottom right corner of your keyboard. 3) release ALT.
This √
dont we have to take the cubed root?
Let's do the problem.
okay
\(\normalsize\color{black}{ \sqrt[3]{5^3 \times 2}+\sqrt[3]{3^3 \times 2 }-\sqrt[3]{2^3 \times 2} }\) I broke it down :)
I would almost imagine something like \[\sum_{n=1}^{5}~~\sqrt[3]{2(6-n)^3}\]
dont i break down the 250?
yes 5³ × 2
make the radican the same bu simplifying?
\(\LARGE \sqrt[3]{250} + \sqrt[3]{54} - \sqrt[3]{16} \) \(\LARGE =\sqrt[3]{125 \times 2} + \sqrt[3]{27 \times 2} - \sqrt[3]{8 \times 2} \) \(\LARGE =\sqrt[3]{125} \times \sqrt[3]{2} + \sqrt[3]{27} \times \sqrt[3]{2} - \sqrt[3]{8} \times \sqrt[3]{2} \) \(\LARGE =5\sqrt[3]{2} + 3 \sqrt[3]{2} - 2\sqrt[3]{2} \) Now you have the same roots.
Rule: \(\Large\color{black}{ \sqrt[a]{b^a}=\sqrt[\cancel{ a }]{b^\cancel{ a }}=b }\)
oh haha! i see!
thank you @MaxwellSmart and @SolomonZelman
yw
anytime
wait? does it need to be simplified?
factor out whats common @MaxwellSmart
:( sorry
Since the roots are all the same, they are now like terms. Now you can combine them together.
combine what though?
Just like 5a + 3a - 2a = 6a
oh...duh you just showed me that
thanks again!
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