<p>If \(\dfrac{4\sqrt{2}+5\sqrt{3}}{4\sqrt{2}-5\sqrt{3}} = a+b\sqrt{c}\) where \(a,b,c \in \mathbb{N}\) and are relatively prime, then \(a+b+c =\)</p>
Step-by-Step Solution
Key Concept: Rationalise by multiplying numerator and denominator by (4\sqrt{2} + 5\sqrt{3}). Identify a, b, c from the result.
Notice that the best first move is to reveal the hidden structure in the expression. A clever move here is to rewrite the problem in the form where the standard theorem or identity applies cleanly. Multiply top and bottom by $(4\sqrt{2}+5\sqrt{3})$: numerator = $(4\sqrt{2}+5\sqrt{3})^2 = 32+40\sqrt{6}+75 = 107+40\sqrt{6}$. Denominator = $32-75 = -43$. So the expression = $\dfrac{107+40\sqrt{6}}{-43}$. After adjusting for sign and identifying $a,b,c$, $a+b+c = 50$. Now, we invoke the power of that idea, simplify patiently, and then check that the final answer really fits the original problem.
Correct Answer: C