JEE Main 2025 — 2 April, Morning Shift — Question 28

$(x+1)=\left[\left(x^{1 / 3}\right)+1\right]\left[x^{2 / 3}-x^{1 / 3}+1\right]$

$(x+1)=(\sqrt{x}-1)(\sqrt{x}+1)$

Now,

$\left(\frac{x+1}{x^{2 / 3}-x^{1 / 3}+1}\right)=\left(x^{1 / 3}+1\right)$

$\frac{x-1}{x-x^{1 / 2}}=\frac{(\sqrt{x}-1)(\sqrt{x}+1)}{(\sqrt{x})^{2}-\sqrt{x}}$

$\left[\left(x^{1 / 3}+1\right)-\left(1+\frac{1}{\sqrt{x}}\right)\right]^{10}=\left[x^{1 / 3}-\frac{1}{x^{1 / 2}}\right]^{10}$

$T_{r+1}={ }^{10} C_{r}\left[-\frac{1}{x^{1 / 2}}\right]^{r} \cdot\left(x^{1 / 3}\right)^{10-r}$

$\Rightarrow \quad x^{\left(\frac{10-r}{3}-\frac{r}{2}\right)} \cdot{ }^{10} C_{r}(-1)^{r}$

The term independent of $x$ when exponent of $x$ is 0 .

$\Rightarrow r=4$

So, term $\rightarrow{ }^{10} C_{4}(-1)^{4} x^{0}=210$