Let t be real number such that `t^2=at+b` for some positive integers a and b. Then forany choice of positive integers a and b.`t^3` is never equal to

Options

A. 4t+3
B. 8t+5
C. 10t+3
D. 6t+5

Answer

b

Explanation

`t^(2)=at+b,a,bin 1^(+)`
`t^(3)=at^(2)+bt`
a(at +b)+bt
`a^(2)t+bt=bt`
`Rightarrow t^(3)= (a^(2)+b) t +ab` , check possibilty for a,b, `in I^(+)` from options.
`a^(2)+b=4`
ab=3 possible
(b) `a^(2)+b=8`
ab=5 not possible
(c) `a^(2) +b=10`
ab = 3 possible
(d) `a^(2) +b=6`
ab=5 possible