Mastering Algebra Challenge- With a Twist!

[@ManojkantSamal]

Ans ::272286..May be
Explain later
^=read as to the power
*=read as square root
Let x=a, b=(1/x)
So, ab=1....eqn1
As per question
a^3+b^3=18.......eqn3
(a+b)^3-{3ab(a+b)=18
(a+b)^3-3(a+b)=18..(ab=1)
Let a+b=R
So,
R^3-3R=18
Though cubic equation so H&T for bagging the first root
Let R=3
R^3-3R=(3^3)-(3×3)
=27-9=18
So R=3
a+b=3......eqn2
Let a+b=R
Now,
a^5+b^5=(a+b)^5-[5ab(a+b){(a+b)^2-ab}]
So,
a^5+b^5=(3^5)-[(5×3){(3^2)-1}]
=243-{15(9-1)}
=243-(15×8)
=243-120
=123
Now,
a^10+b^10 =(a^5+b^5)^2-(2×a^5×b^5)
=(123)^2 - 2
=15129 -2=15127
So,
(X^10)+(1/x^10)=15127......eqn4
Eqn3 ×eqn4
{X^3+(1/x^3)}{x^10+(1/x^10)}=15127×18
So,
{X^3×(x^10)}+{x^3×(1/x^10)}+{(1/x3)×x^10}+{(1/x^3)×(1/x^10)}=272286
So,
(X^13)+(1/x^7)+(x^7)+(1/x^13)=272286
So,
(X^7)+(1/x^7)+(x^13)+(1/x^13)=272286......

[@Fjfurufjdfjd]

Ε=272286
χ+1/χ=3 , χ^2+1/χ^2=7 , χ^3+1/χ^3=18 , χ^4+1/χ^4=47 , χ^7+1/χ^7=843 , χ^6+1/χ^6=322 , χ^13+1/χ^13=271443
Ε=843+271443=272286

[@Shobhamaths]

I too got same answer 272286

[@nasrullahhusnan2289]

x³+(1/x³)=18 --> x>0
[x+(1/x)]³=x³+(1/x³)+3[x+(1/x)]
=18+3[x+(1/x)]
[x+(1/x)][{x+(1/x)}²-3]=3(3²-3)
x+(1/x)=3
x²+(1/x²)=[x+(1/x)]²-2
=7
x³+(1/x³)=18
x⁴+(1/x⁴)=[x²+(1/x²)²-2
=47
x⁷+(1/x⁷)=[x³+(1/x³)][x⁴+(1/x⁴)]
-[x+(1/x)]
=18(47)-3
x⁵+(1/x⁵)=[x²+(1/x²)][x³+(1/x³)]
-[x+(1/x)]
=7(18)-3
x⁶+(1/x⁶)=[x³+(1/x³)]²-2
=18²-2
x¹³+(1/x¹³)=[x⁶+(1/x⁶)][x⁷+(1/x⁷)]
-[x+(1/x)]

[@gregevgeni1864]

272286

[@Quest3669]

X+1/x= 3; x^6+(1/x)^6= 322; x^7+(1/x)^7=843 hence
?= 271443+843= 272286 soln

[@neerajgupta7856]

272286