Given equation: $\frac{a^3+3a}{3a^2+1}=\frac{91}{37}$

To find: The value of 'a'

Solution:

To solve the given equation, we will follow the steps below:

Step 1: Cross-multiplication

$\Rightarrow (a^3+3a) \cdot 37 = (3a^2+1) \cdot 91$

Step 2: Expand the equation

$\Rightarrow 37a^3 + 111a = 273a^2 + 91$

Step 3: Rearrange the equation

$\Rightarrow 37a^3 - 273a^2 + 111a - 91 = 0$

Step 4: Factorize the equation

$\Rightarrow (37a^3 - 273a^2) + (111a - 91) = 0$

$\Rightarrow 37a^2(a - 7) + 111(a - 7) = 0$

$\Rightarrow (a - 7)(37a^2 + 111) = 0$

Step 5: Set each factor equal to zero and solve for 'a'

Case 1: $a - 7 = 0$

$\Rightarrow a = 7$

Case 2: $37a^2 + 111 = 0$

$\Rightarrow 37a^2 = -111$

$\Rightarrow a^2 = \frac{-111}{37}$

Taking the square root on both sides,

$\Rightarrow a = \pm \sqrt{\frac{-111}{37}}$

However, since we are dealing with real numbers, the value of 'a' cannot be the square root of a negative number. Therefore, we discard the negative square root.

Final Answer: 'a' equals 7.

Therefore, the value of 'a' in the given equation is 7.