$f(x+h)=0$ represents the transformed equation of the equation $f(x)=x^4+2x^3-19x^2-8x+60=0$. If this transformation removes the term containing $x^3$ from $f(x)=0$,then $h=$

Using the Binomial Theorem,evaluate $(99)^{5}$.

If the sum of the coefficients in the expansion of $(\alpha x^2 - 2x + 1)^{35}$ is equal to the sum of the coefficients in the expansion of $(x - \alpha y)^{35}$,then $\alpha = $

Let $\alpha, \beta, \gamma$ and $\delta$ be the coefficients of $x^7, x^5, x^3$ and $x$ respectively in the expansion of $(x+\sqrt{x^3-1})^5+(x-\sqrt{x^3-1})^5, x>1$. If $u$ and $v$ satisfy the equations $\alpha u+\beta v=18$ and $\gamma u+\delta v=20$,then $u+v$ equals:

Evaluate the expression: $\frac{1}{81^{n}} - {}^{2n}C_1 \frac{10}{81^{n}} + {}^{2n}C_2 \frac{10^2}{81^{n}} - \dots + \frac{10^{2n}}{81^{n}} = $

If $x$ is so small that $x^3$ and higher powers of $x$ may be neglected,then $\frac{(1 + x)^{3/2} - (1 + \frac{1}{2}x)^3}{(1 - x)^{1/2}}$ may be approximated as