Math, asked by Itzraisingstar, 7 months ago

If (a+1/a)^2=3 and a>0,prove that a^3+1/a^3=0.

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Answered by Anonymous

Answer:

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a^3 +b^3 = a^2 - ab + b^2

(a^3+ 1/a^3) = ( a+ 1/a) ( a^2 -1 +1/a^2) (1) equation

it is given that

(a+1/a)^2 = 3

( a+ 1/a) = √3. (2) equation

Also

( a+ 1/a)^2 = 3

or - a^2 + 1/a^2 +2 =3

then ( transpose 2 to subtract 3)

a^2 + 1/a^2 = 3-2

a^2 + 1/a^2 = 1. (3) equation

putting 2 and 3 in 1 we have

a^3 +1/a^3 = √3 (1-1) = 0

so the answer is 0

hope it helps

thank you

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Answered by Anonymous

Answer:

$a^3 +b^3 = a^2 - ab + b^2(a^3+ 1/a^3) = ( a+ 1/a) ( a^2 -1 +1/a^2) (1) equationit is given that(a+1/a)^2 = 3or( a+ 1/a) = √3. (2) equationAlso( a+ 1/a)^2 = 3o^2 + 1/a^2 = 3-2a^2 + 1/a^2 = 1. (3) equationputting 2 and 3 in 1 we havea^3 +1/a^3 = √3 (1-1) = 0so the answer is 0Follow me ✌️$

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If (a+1/a)^2=3 and a>0,prove that a^3+1/a^3=0. No spams.

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If (a+1/a)^2=3 and a>0,prove that a^3+1/a^3=0.

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