Let x = √72/[√2(√13+1)] = 6/(√13+1) = 1/2(√13-1). So, x^3=2√13-5, x^6=77-20√13 and x^12 = 11,129-3080√13 = a +b√13. Thus, a =11,129 and b = -3080 > a+3b=1889.
*=read as square root ^=read as to the power *72/(*26+*2) =(6.*2)/{*2(*13+1)} =6/(*13+1)={6(*13-1)}/12 (multiply both numerator &denominator by *13-1) =(*13-1)/2.......eqn1 Squaring eqn1 {(13+1)-2.*13}/4 =2(7-*13)/4=(7-*13)/2...eqn2 Squaring eqn2 {(49+13)-14.*13}/4 =2(31-7.*13)/4 =(31-7.*13)/2.....eqm3 Take the cube of eqn3 Numerator (31)^3-(7.*13)^3+3.317.*13(7.*13-31) =29791-4459.*13+59241-20181.*13 =89032-24640.*13 =8(11129-3080.*13) Denominator 2^3=8 So, 8(11129-3080.*13)/8 =11129-3080.*13 As per question a+(b.*13)=11129-3080.*13 So, a=11129 b.*13=-3080.*13 b=-3080 So a+3b=11129+{3×(-3080)} =11129-9240 =789
Let x = √72/[√2(√13+1)] = 6/(√13+1) = 1/2(√13-1). So, x^3=2√13-5, x^6=77-20√13 and x^12 = 11,129-3080√13 = a +b√13. Thus, a =11,129 and b = -3080 > a+3b=1889.
*=read as square root
^=read as to the power
*72/(*26+*2)
=(6.*2)/{*2(*13+1)}
=6/(*13+1)={6(*13-1)}/12 (multiply both numerator &denominator by *13-1)
=(*13-1)/2.......eqn1
Squaring eqn1
{(13+1)-2.*13}/4
=2(7-*13)/4=(7-*13)/2...eqn2
Squaring eqn2
{(49+13)-14.*13}/4
=2(31-7.*13)/4
=(31-7.*13)/2.....eqm3
Take the cube of eqn3
Numerator
(31)^3-(7.*13)^3+3.317.*13(7.*13-31)
=29791-4459.*13+59241-20181.*13
=89032-24640.*13
=8(11129-3080.*13)
Denominator
2^3=8
So,
8(11129-3080.*13)/8
=11129-3080.*13
As per question
a+(b.*13)=11129-3080.*13
So,
a=11129
b.*13=-3080.*13
b=-3080
So
a+3b=11129+{3×(-3080)}
=11129-9240
=789
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