Question

If $a=4,b=0$and $c=-1$, find the value of $\frac{2a^{3}b+3b^2{c}^2-ac}{a^2+3ab}$.

Answer 1

Here we have to find the value of expression for given value of $a,b,\&c$

i.e- $\frac{2a^{3}b+3b^2{c}^2-ac}{a^2+3ab}$

Putting the value of $a=4,b=0$and $c=-1$

we have,

$$\begin{gathered} \frac{2\times {\left(4\right)}^3\times 0+3\times 0^2\times {(-1)}^2-4\times (-1)}{4^2+3\times 4\times 0} \\ =\frac{2\times 64\times 0+3\times 0\times 1+4}{16+12\times 0} \\ =\frac{0+0+4}{16+0} \\ =\frac{4}{16} \\ =\frac{1}{4} \end{gathered}$$

Hence, the value of the expression is $\frac{1}{4}$