1 If a = p^1/3-p^-1/3
prove that: a^3 + 3a = p - 1/p

angeline2004   ·   12.08.2020 04:01
ANSWER(S): 1 Show answers 8 Сomment
ANSWER(S)
answered: krazyykay
28.06.2019 13:40

p = 6s

step-by-step explanation:

answered: savdeco
28.06.2019 10:00

step-by-step explanation: s

answered: xcji7384
26.06.2019 21:20

a

step-by-step explanation:

when a number is between two straight lines it means the absolute value.

to find absolute value you need to know how many numbers is that number away from 0.

when finding absolute value the number will always come out positive

so absolute value of -3 is 3

absolute value of 3/2 is 3/2

absolute value of -2 is 2

absolute value of 3.5 is 3.5

absolute value of -1 is 1

now put these numbers in order:

a)   |-1|, |3/2|, |-2|, |-3|, |3.5|

answered: annyarias3563
30.01.2022 20:25

Hello, please consider the following.

We know that

a = p^{\frac{1}{3}}-p^{-\frac{1}{3}}\\\\=p^{\frac{1}{3}}-\dfrac{1}{p^{\frac{1}{3}}}

And we can write that.

(p-\dfrac{1}{p})^3=(p-\dfrac{1}{p})(p^2-2+\dfrac{1}{p^2})\\\\=p^3-2p+\dfrac{1}{p}-p+\dfrac{2}{p}-\dfrac{1}{p^3}\\\\=p^3-\dfrac{1}{p^3}-3(p-\dfrac{1}{p})

It means that, by replacing p by p^{1/3}

(p^{1/3}-\dfrac{1}{p^{1/3}})^3=p-\dfrac{1}{p}-3(p^{1/3}-\dfrac{1}{p^{1/3}})\\\\\\\text{ So }\\\\a^3=p-\dfrac{1}{p}-3a\\\\\boxed{ a^3+3a=p-\dfrac{1}{p} }

Hope this helps.

Do not hesitate if you need further explanation.

Thank you

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