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Từ $\frac ab=\frac cd$ ta có $ad=bc$. Ta có $(a^{24}+b^{24})(c^{25}-d^{25})=(a^{24}+b^{24})c^{25}-(a^{24}+b^{24})d^{25}=a^{24}c^{25}+b^{24}c^{25}-a^{24}d^{25}-b^{24}d^{25}=a^{24}c^{25}-a^{24}d^{25}+b^{24}c^{25}-b^{24}d^{25}=a^{24}(c^{25}-d^{25})+b^{24}(c^{25}-d^{25})=(a^{24}+b^{24})(c^{25}-d^{25})$. Tương tự $(c^{24}+d^{24})(a^{25}-b^{25})=(c^{24}+d^{24})a^{25}-(c^{24}+d^{24})b^{25}=c^{24}a^{25}+d^{24}a^{25}-c^{24}b^{25}-d^{24}b^{25}=c^{24}a^{25}-c^{24}b^{25}+d^{24}a^{25}-d^{24}b^{25}=c^{24}(a^{25}-b^{25})+d^{24}(a^{25}-b^{25})=(c^{24}+d^{24})(a^{25}-b^{25})$. Do đó $\frac{(a^{24}+b^{24})^{25}}{(c^{24}+d^{24})^{25}}=\frac{(a^{25}-b^{25})^{24}}{(c^{25}-d^{25})^{24}}$.