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Unless $ f(x) - 10x $ has four zeros — impossible for degree 3. So the only way the conditions hold is if $ f(x) = 10x $, and accept its not truly cubic — but the problem states it is.
Wait — perhaps the function is cubic but satisfies these values — but four points determine a cubic fully only if conditions match. But three points determine a quadratic, four dont restrict to a cubic unless consistent.
But here: $ f(1)=10, f(2)=20, f(3)=30, f(4)=40 $ — these lie on $ f(x)=10x $. Any cubic passing through these four points must satisfy $ f(x) = 10x + (x-1)(x-2)(x-3)(x-4)q(x) $, but degree issue.
$ x=4 $: $ 64a + 16b + 4c + d = 40 $
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