need answer for this question 2nd one.
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since a is a root of the equation x^2-3x+1=0
we have a^2-3a+1=0
Or, a-3+1/a=0(dividing both sides by a)
Or,a+1/a=3
Now a^3/a^6+1
=1/(a^3+1/a^3)(dividing numerator and denominator by a^3)
=1/{(a+1/a)^3-3ĆaĆ1/a(a+1/a)}
=1/{(3)^3-3(3)}
=1/(27-9)
=1/18
we have a^2-3a+1=0
Or, a-3+1/a=0(dividing both sides by a)
Or,a+1/a=3
Now a^3/a^6+1
=1/(a^3+1/a^3)(dividing numerator and denominator by a^3)
=1/{(a+1/a)^3-3ĆaĆ1/a(a+1/a)}
=1/{(3)^3-3(3)}
=1/(27-9)
=1/18
popstar12:
hi
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