Detailed Solution:
We are given:
\(\sqrt[3]{x} + \sqrt[3]{y} = 13 \quad \text{(1)}\)
\(\sqrt{x} + \sqrt{y} = 35 \quad \text{(2)}\)
Let \( a = \sqrt[6]{x} \) and \( b = \sqrt[6]{y} \). Then:
- \( \sqrt[3]{x} = a^2 \), \( \sqrt[3]{y} = b^2 \)
- \( \sqrt{x} = a^3 \), \( \sqrt{y} = b^3 \)
Substituting into (1) and (2):
\( a^2 + b^2 = 13 \quad \text{(3)} \)
\( a^3 + b^3 = 35 \quad \text{(4)} \)
Use the identity:
\( a^3 + b^3 = (a + b)^3 – 3ab(a + b) \)
\( a^2 + b^2 = (a + b)^2 – 2ab \)
Let \( S = a + b \), \( P = ab \). Then:
- \( S^2 – 2P = 13 \Rightarrow P = \frac{S^2 – 13}{2} \)
- \( a^3 + b^3 = S^3 – 3PS = 35 \)
Substitute:
\( S^3 – \frac{3(S^2 – 13)}{2}S = 35 \)
\( 2S^3 – 3S^3 + 39S = 70 \Rightarrow -S^3 + 39S = 70 \Rightarrow S^3 – 39S + 70 = 0 \)
Solving:
\( (S – 2)(S + 7)(S – 5) = 0 \Rightarrow S = 2, -7, 5 \)
Try \( S = 5 \):
\( P = \frac{25 – 13}{2} = 6 \)
\( a + b = 5, ab = 6 \Rightarrow (t – 2)(t – 3) = 0 \Rightarrow a = 2, b = 3 \)
Then:
- \( x = a^6 = 2^6 = 64 \)
- \( y = b^6 = 3^6 = 729 \)
Final Answer:
\(\boxed{x = 64,\ y = 729} \quad \text{or} \quad \boxed{x = 729,\ y = 64}\)