+26387 xpand by using Binomial theorem (5-x)^6
\(\begin{array}{|rcll|} \hline && (5-x)^6 \\ &=& [(-1)\cdot (x-5)]^6 \\ &=& (-1)^6 \cdot (x-5)^6 \quad & | \quad (-1)^6 = 1 \\ &=& (x-5)^6 \\ \hline \end{array} \)
\(\begin{array}{|rcll|} \hline && (x-5)^6 \\ &=& \binom60 x^6\cdot 5^0 - \binom61 x^5\cdot 5^1 + \binom62 x^4\cdot 5^2 - \binom63 x^3\cdot 5^3 \\ && + \binom64 x^2\cdot 5^4 - \binom65 x^1\cdot 5^5 + \binom66 x^0\cdot 5^6 \\ &=& \binom60 x^6 - 5^1\cdot \binom61 x^5 + 5^2\cdot \binom62 x^4 - 5^3\cdot \binom63 x^3 \\ && + 5^4\cdot \binom64 x^2 - 5^5\cdot \binom65 x^1 + 5^6 \cdot \binom66\\ && \begin{array}{|rcll|} \hline \binom60 = \binom66 = 1 \\ \binom61 = \binom65 = 6 \\ \binom62 = \binom64 = 15 \\ \binom63 = 20 \\ \hline \end{array} \\ &=& 1\cdot x^6 - 5^1\cdot 6 \cdot x^5 + 5^2\cdot 15 \cdot x^4 - 5^3\cdot 20 \cdot x^3 \\ && + 5^4\cdot 15\cdot x^2 - 5^5\cdot 6 \cdot x^1 + 5^6 \cdot 1\\ &=& x^6 - 30\ x^5 +375\ x^4-2500\ x^3 + 9375\ x^2-18750\ x+15625 \\ \hline \end{array} \)
