If \(\mathrm{a}=\frac{\mathrm{b}^{2}}{\mathrm{~b}-\mathrm{a}}\) then the value of \(\mathrm{a}^{3}+\mathrm{b}^{3}\) is (a) 2 (b) \(6 \mathrm{ab}\) (c) 0 (d) 1

Given the equation \(a = \frac{b^2}{b-a}\), we can rearrange this equation to find a relationship between \(a\) and \(b\). Multiplying both sides by \(b-a\) gives:

\[
a(b – a) = b^2
\]

Expanding the left side:

\[
ab – a^2 = b^2
\]

Rearranging terms:

\[
ab = a^2 + b^2
\]

Now, we are asked to find the value of \(a^3 + b^3\). We know the identity for the sum of cubes is:

\[
a^3 + b^3 = (a + b)(a^2 – ab + b^2)
\]

Substituting \(ab = a^2 + b^2\) into the identity:

\[
a^3 + b^3 = (a + b)(0)
\]

Since anything multiplied by 0 is 0:

\[
a^3 + b^3 = 0
\]

Therefore, the value of \(a^3 + b^3\) is \(0\), and the correct option is:

(c) 0