Write down the first three terms in the expansion, in descending powers of x, of [ax-(b/x^3)]^8. Hence evaluate the term independent of xof [5x-(1/x^3)]^8.

I dunno how to do the Hence part, making use of the terms in the expansion.

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- Oct 8th 2008, 09:00 AMmaybeline9216binomial thm
**Write down the first three terms in the expansion, in descending powers of x, of [ax-(b/x^3)]^8. Hence evaluate the term independent of x****of [5x-(1/x^3)]^8.**

I dunno how to do the Hence part, making use of the terms in the expansion. - Oct 8th 2008, 09:56 AMThePerfectHacker
In this expansion we have terms of form: $\displaystyle {8\choose k} (-1)^k(5x)^k \left( \frac{1}{x^3} \right)^{8-k}$

To have a term independent of $\displaystyle x$ we require that the $\displaystyle x$ in numerator cancels out the $\displaystyle x$ in denominator. The exponent of $\displaystyle x$ in numberator is $\displaystyle k$ and the exponent of $\displaystyle x$ in denominator is $\displaystyle 3(8-k)$. We want that $\displaystyle k = 3(8-k)$ so that they cancel out. Therefore, $\displaystyle k=6$ if we solve this equation.

Therefore the indepentent term is: $\displaystyle {8\choose 6} (-1)^k 5^6$ - Oct 8th 2008, 08:04 PMmaybeline9216