JEE Main 2026 — 6 April, Evening Shift — Question 41

$(1-x)^{3}=1-x^{3}-3 x(1-x)$ $(1-x)^{3}=1-x^{3}-3 x+3 x^{2}$ $(1-x)^{3}+3 x-3 x^{2}=1-x^{3}$ $\left(1-x^{3}\right)^{10}=\left[(1-x)^{3}+3 x(1-x)\right]^{10}$ $\Rightarrow{ }^{10} \mathrm{C}_{\mathrm{r}}(3 \mathrm{x}(1-\mathrm{x}))^{\mathrm{r}} \cdot\left((1-\mathrm{x})^{3}\right)^{10-\mathrm{r}}$ $\left(1-\mathrm{x}^{3}\right)^{10}={ }^{10} \mathrm{C}_{\mathrm{r}} \cdot 3^{\mathrm{r}} \cdot \mathrm{x}^{\mathrm{r}} \cdot(1-\mathrm{x})^{30-2 \mathrm{r}}$ $\therefore \mathrm{a}_{\mathrm{r}}={ }^{10} \mathrm{C}_{\mathrm{r}} \cdot 3{ }^{\mathrm{r}}$ $\frac{\mathrm{a}_{9}}{\mathrm{a}_{10}}=\frac{{ }^{10} \mathrm{C}_{9} 3^{9}}{{ }^{10} \mathrm{C}_{10} .3^{10}}$ $\frac{\mathrm{a}_{9}}{\mathrm{a}_{10}}=\frac{10}{3}$ $\therefore \frac{9 \mathrm{a}_{9}}{\mathrm{a}_{10}}=30$