An algebra problem by Saksham Mittal

@Saksham Mittal Make the problem better by changing it into latex.There are many guides which may help.And as for your question"Is it difficult???" Nope it's not

Let $\sqrt{a}=c$ and $\sqrt{b}=d$ so that $a+b=c^2+d^2$ .

Now $c^3+d^3=183$ and $c^2d+cd^2=cd(c+d)=182$ .

Note that $c^3+d^3=(c+d)^3-3cd(c+d)=183$ $\implies$ $(c+d)^3-3 \times 182 = 183$ .

$\implies (c+d)^3=729 \implies c+d=9$ and therefore $cd=\frac{182}{9}$ .

Now

$\begin{aligned} a+b & = c^2+d^2 = (c+d)^2-2cd \\ & = 81 - \frac{364}{9} \\ & = \frac{365}{9} \end{aligned}$

Therefore $\frac95(a+b)=\frac95 \times \frac{365}9=\boxed{73}$

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