^= read as to the power *=read as square root As per question b+b^2 +b^3 +(1/b)+(1/b^2)+(1/b^3)=6 {b+(1/b)}+{b^2+(1/b^2)}+{b^3+(1/b^3)}=6 Let b+(1/b)=X So X+{b^2+(1/b^2)}+{b^3+(1/b^3)} Now explain b^2+(1/b^2) ={b+(1/b)}^2 -2.b. (1/b) =x^2 -2 The explanation of {b^3+(1/b^3) will be as below b^3+(1/b^3) ={b+(1/b)}^3-3.b. (1/b){b+(1/b)} =x^3 -3x Now the given equation will be X+x^2-2+x^3-3x=6 =x^3+x^2-2x=8 Though cubic equation so the first root will be found out by Hit & Trial method X^3+x^2-2x (2^3)+(2^2) -(2×2) =8+4-4=8 So x=2 Implies {b+(1/b)}=2 {b^2+1}/b=2 b^2+1=2b b^2-2b+1=0 Here a=1, b=(-2), c=1 D=b^2-4ac =(-2)^2-{4×1×1} =4-4=0 b={-b±*(D)}/2a ={-(-2)±*(0)}/2 =(2±0)/2 =2/2 =1 Hence b=1
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^= read as to the power
*=read as square root
As per question
b+b^2 +b^3 +(1/b)+(1/b^2)+(1/b^3)=6
{b+(1/b)}+{b^2+(1/b^2)}+{b^3+(1/b^3)}=6
Let b+(1/b)=X
So
X+{b^2+(1/b^2)}+{b^3+(1/b^3)}
Now explain
b^2+(1/b^2)
={b+(1/b)}^2 -2.b. (1/b)
=x^2 -2
The explanation of {b^3+(1/b^3) will be as below
b^3+(1/b^3)
={b+(1/b)}^3-3.b. (1/b){b+(1/b)}
=x^3 -3x
Now the given equation will be
X+x^2-2+x^3-3x=6
=x^3+x^2-2x=8
Though cubic equation so the first root will be found out by Hit & Trial method
X^3+x^2-2x
(2^3)+(2^2) -(2×2)
=8+4-4=8
So x=2
Implies {b+(1/b)}=2
{b^2+1}/b=2
b^2+1=2b
b^2-2b+1=0
Here a=1, b=(-2), c=1
D=b^2-4ac
=(-2)^2-{4×1×1}
=4-4=0
b={-b±*(D)}/2a
={-(-2)±*(0)}/2
=(2±0)/2
=2/2 =1
Hence b=1
I am glad that I have reviewed how to improve my substitution abilities. And the answer is b = 2.