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Question Number 170945 by Rasheed.Sindhi last updated on 04/Jun/22

y=(9)^(1/3) +(3)^(1/3) +4 ; (1+(1/y))^3 =?

y=93+33+4;(1+1y)3=?

Answered by cortano1 last updated on 04/Jun/22

y−3=(9)^(1/3) +(3)^(1/3) +1 (y−3)((3)^(1/3) −1)=2 y=(2/( (3)^(1/3) −1))+3=((3(3)^(1/3) −1)/( (3)^(1/3) −1)) (1+(1/y))^3 =(1+(((3)^(1/3) −1)/(3(3)^(1/3) −1)))^3 =(((4(3)^(1/3) −2)/(3(3)^(1/3) −1)))^3 =((48(3)^(1/3) −96(9)^(1/3) +184)/(9(3)^(1/3) −27(9)^(1/3) +80))

y−3=93+33+1(y−3)(33−1)=2y=233−1+3=333−133−1(1+1y)3=(1+33−1333−1)3=(433−2333−1)3=4833−9693+184933−2793+80

Commented by Rasheed.Sindhi last updated on 06/Jun/22

Thank you sir!

Thankyousir!

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