Let $a$ and $b$ be real numbers such that $a+b=7$ and $a b=5$. What is the value of $a^{3}+b^{3}$? We will not seek to explicitly determine $a$ and $b$.

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The problem can be solved using the identity for the sum of cubes:
\[ a^3 + b^3 = (a+b)(a^2 - ab + b^2) \]

First, we need to find \(a^2 + b^2\). We can use the square of the sum of \(a\) and \(b\):
\[ (a+b)^2 = a^2 + 2ab + b^2 \]

Given \(a + b = 7\) and \(ab = 5\), we substitute these values into the equation:
\[ 7^2 = a^2 + 2 \cdot 5 + b^2 \]
\[ 49 = a^2 + 10 + b^2 \]
\[ a^2 + b^2 = 49 - 10 \]
\[ a^2 + b^2 = 39 \]

Next, we substitute \(a^2 + b^2\) and \(ab\) into the identity for the sum of cubes:
\[ a^3 + b^3 = (a+b)(a^2 - ab + b^2) \]
\[ a^3 + b^3 = 7(a^2 + b^2 - ab) \]
\[ a^3 + b^3 = 7(39 - 5) \]
\[ a^3 + b^3 = 7 \cdot 34 \]
\[ a^3 + b^3 = 238 \]

Thus, the value of \(a^3 + b^3\) is \(\boxed{238}\).
On a $(a+b)^{3}=a^{3}+3 a^{2} b+3 a b^{2}+b^{3}=a^{3}+b^{3}+3 a b(a+b)$, so $a^{3}+b^{3}=(a+b)^{3}-3 a b(a+b)=$ $7^{3}-3 \cdot 5 \cdot 7=7(49-15)=7 \times 34=238$.