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Last edited by oceanmd; Dec 13th 2007 at 05:04 AM.
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Originally Posted by oceanmd $\displaystyle (x^2-y)^6$ $\displaystyle {6\choose3}(x^2)^3(-y)^0=20x^6$ Wil you please check the solution and correct if it is wrong. Thank you why is the power of -y zero?
Originally Posted by oceanmd $\displaystyle {6\choose3}(x^2)^3(-y)^3=20x^6(-y)^3$ Is it correct now? Thanks yes, using of course, $\displaystyle (a + b)^n = \sum_{k = 0}^n {n \choose k}a^kb^{n - k}$ write it as $\displaystyle -20x^6y^3$
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