If $a+b+c=6, a^{2}+b^{2}+c^{2}=14$ and $a^{3}+b^{3}+c^{3}=36,$ then the value of $abc$ is

$4 \frac{3}{4} + 2 \frac{1}{8} + 7 \frac{1}{4} + 3 \frac{7}{8} + 11 \frac{12}{13} = ?$

$\frac{3}{4} \left(1+\frac{1}{3}\right) \left(1+\frac{2}{3}\right) \left(1-\frac{2}{5}\right) \left(1+\frac{6}{7}\right) \left(1-\frac{12}{13}\right)$ is equal to

If $1$ is added to the denominator of a fraction,the fraction becomes $\frac{1}{2}$. If $1$ is added to the numerator of the fraction,the fraction becomes $1$. The fraction is

If $4x + 5y = 83$ and $\frac{3x}{2y} = \frac{21}{22}$,then $y - x = ?$

If $a=25, b=15, c=-10$; then the value of $\frac{a^{3}+b^{3}+c^{3}-3abc}{(a-b)^{2}+(b-c)^{2}+(c-a)^{2}}$ is