^=read as to the power *=read as square root As per question a^(1/6)={a^(1/3)+1}/18 {a^(1/3) +1}/{a^(1/6)} =18 {a^(1/3)/a^(1/6)}+{1/a^(1/6)}=18 [a^{(1/3)-(1/6)}]+{1/a^(1/6)}=18 a^(1/6)+{1/a^(1/6)=18 Squaring the equation a^(1/3)+{1/a^(1/3)+2=324 a^(1/3)+{1/a^(1/3)=324-2=322 {a^(1/9)}^3+{1/a^(1/9)}^3=322 Let {a^(1/9)}=x So, X^3+(1/x^3 )=322 {X+(1/x)}^3 -{3.x. (1/x)}{x^3+(1/x^3)=322 So, {X+(1/x)^3-3{x+(1/x)}=322 Let {x+(1/x)}=R So, R^3-3R=322 R^3-3R-322=0 Though cubic equation mostly it takes the implimentation of 'Hit &Trial' method to discover it's first root So, If R=7, then R^3-3R-322=0 (7)^3-(3×7)-322=343-21-322 =343-343=0, which satisfies the equation So R=7 R= {x+(1/x)}=[a^(1/9)+{1/a^(1/9)}] Hence {a^(1/9)}+{1/a^(1/9)}=7
Sir,...your solution is very interesting, but I have a question...Where did you learn to write the symbol of the root, so ugly. If you don't write it normally... like millions of people? ...I will never look at your solutions.Capisci?
Good question
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The answer is 7 and I shall use that for practice!!!
^=read as to the power
*=read as square root
As per question
a^(1/6)={a^(1/3)+1}/18
{a^(1/3) +1}/{a^(1/6)} =18
{a^(1/3)/a^(1/6)}+{1/a^(1/6)}=18
[a^{(1/3)-(1/6)}]+{1/a^(1/6)}=18
a^(1/6)+{1/a^(1/6)=18
Squaring the equation
a^(1/3)+{1/a^(1/3)+2=324
a^(1/3)+{1/a^(1/3)=324-2=322
{a^(1/9)}^3+{1/a^(1/9)}^3=322
Let {a^(1/9)}=x
So,
X^3+(1/x^3 )=322
{X+(1/x)}^3 -{3.x. (1/x)}{x^3+(1/x^3)=322
So,
{X+(1/x)^3-3{x+(1/x)}=322
Let {x+(1/x)}=R
So,
R^3-3R=322
R^3-3R-322=0
Though cubic equation mostly it takes the implimentation of 'Hit &Trial' method to discover it's first root So,
If R=7, then
R^3-3R-322=0
(7)^3-(3×7)-322=343-21-322
=343-343=0, which satisfies the equation
So R=7
R= {x+(1/x)}=[a^(1/9)+{1/a^(1/9)}]
Hence
{a^(1/9)}+{1/a^(1/9)}=7
Sir,...your solution is very interesting, but I have a question...Where did you learn to write the symbol of the root, so ugly. If you don't write it normally... like millions of people? ...I will never look at your solutions.Capisci?