与えられた式を簡略化する問題です。式は以下の通りです。 $\frac{a(r^{10}-1)}{r-1} \times \frac{r-1}{a(r^{30}-1)}$代数学式の簡略化分数式因数分解累乗2025/8/31. 問題の内容与えられた式を簡略化する問題です。式は以下の通りです。a(r10−1)r−1×r−1a(r30−1)\frac{a(r^{10}-1)}{r-1} \times \frac{r-1}{a(r^{30}-1)}r−1a(r10−1)×a(r30−1)r−12. 解き方の手順まず、式を整理します。a(r10−1)r−1×r−1a(r30−1)=a(r10−1)(r−1)(r−1)a(r30−1)\frac{a(r^{10}-1)}{r-1} \times \frac{r-1}{a(r^{30}-1)} = \frac{a(r^{10}-1)(r-1)}{(r-1)a(r^{30}-1)}r−1a(r10−1)×a(r30−1)r−1=(r−1)a(r30−1)a(r10−1)(r−1)次に、aaaと(r−1)(r-1)(r−1)を約分します。a(r10−1)(r−1)(r−1)a(r30−1)=r10−1r30−1\frac{a(r^{10}-1)(r-1)}{(r-1)a(r^{30}-1)} = \frac{r^{10}-1}{r^{30}-1}(r−1)a(r30−1)a(r10−1)(r−1)=r30−1r10−1r30−1r^{30} - 1r30−1 を (r10)3−1(r^{10})^3 - 1(r10)3−1 と見ると、x3−1=(x−1)(x2+x+1)x^3-1 = (x-1)(x^2+x+1)x3−1=(x−1)(x2+x+1)の公式が使えます。 ここで、x=r10x = r^{10}x=r10とすると、r30−1=(r10)3−1=(r10−1)((r10)2+r10+1)=(r10−1)(r20+r10+1)r^{30} - 1 = (r^{10})^3 - 1 = (r^{10} - 1)((r^{10})^2 + r^{10} + 1) = (r^{10} - 1)(r^{20} + r^{10} + 1)r30−1=(r10)3−1=(r10−1)((r10)2+r10+1)=(r10−1)(r20+r10+1)したがって、r10−1r30−1=r10−1(r10−1)(r20+r10+1)=1r20+r10+1\frac{r^{10}-1}{r^{30}-1} = \frac{r^{10}-1}{(r^{10}-1)(r^{20}+r^{10}+1)} = \frac{1}{r^{20}+r^{10}+1}r30−1r10−1=(r10−1)(r20+r10+1)r10−1=r20+r10+113. 最終的な答え1r20+r10+1\frac{1}{r^{20}+r^{10}+1}r20+r10+11