Find, in ascending powers of \(x\), the first four terms in the expansion of \(\left(\dfrac13-bx\right)^6\), where \(b\) is a constant. Simplify the coefficients in your expansion.[4]
Given that the expansion of \(\left(1+\dfrac3x+\dfrac3{x^2}\right)\left(\dfrac13-bx\right)^6\) has no constant term, find the possible values of \(b\).[4]
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Answer:(a) \(\dfrac1{729}-\dfrac2{81}bx+\dfrac5{27}b^2x^2-\dfrac{20}{27}b^3x^3+\cdots\). (b) \(b=\dfrac19\) or \(\dfrac1{45}\).