1. Given data

We have a cubic polynomial $f(x)$ such that $f(k)=-\frac{2}{k}\quad \text{for } k=2,3,4,5.$

So, $f(2)=-1,\quad f(3)=-\frac23,\quad f(4)=-\frac12,\quad f(5)=-\frac25.$

We need to find: $52-10f(10).$

2. Construct a polynomial with known zeros

Define $g(x)=x f(x)+2.$

Since $f(x)$ is of degree $3$, $x f(x)$ is of degree $4$, so $g(x)$ is a polynomial of degree at most $4$.

Now for $x=2,3,4,5$, $g(k)=k f(k)+2 = k\left(-\frac{2}{k}\right)+2=-2+2=0.$

Thus $g(x)$ has zeros at $x=2,3,4,5$.

Therefore, $g(x)=\lambda (x-2)(x-3)(x-4)(x-5)$ for some constant $\lambda$.

3. Use the fact that $f(x)$ is a polynomial

We have $x f(x)+2 = \lambda (x-2)(x-3)(x-4)(x-5).$

So, $f(x)=\frac{\lambda (x-2)(x-3)(x-4)(x-5)-2}{x}.$

Since $f(x)$ is a polynomial, the numerator must be divisible by $x$.

Hence, substituting $x=0$ into the numerator gives zero: $\lambda (-2)(-3)(-4)(-5)-2=0.$

Now, $(-2)(-3)(-4)(-5)=120.$

So, $120\lambda -2=0 \implies \lambda=\frac{1}{60}.$

Thus, $x f(x)+2=\frac{1}{60}(x-2)(x-3)(x-4)(x-5).$

4. Find $f(10)$

Substitute $x=10$: $10f(10)+2=\frac{1}{60}(10-2)(10-3)(10-4)(10-5).$

That is, $10f(10)+2=\frac{1}{60}(8)(7)(6)(5).$

Compute: $(8)(7)(6)(5)=1680.$

Hence, $10f(10)+2=\frac{1680}{60}=28.$

So, $10f(10)=26.$

Therefore, $f(10)=\frac{26}{10}=\frac{13}{5}.$

5. Compute the required value

We need $52-10f(10)=52-26=26.$

6. Comparison with stored answer

Our derived answer is $26$, which matches the stored correct answer.