If x = (√7 − √5) / (√5 + √7) and y is the reciprocal of x, then what is the…

Concept. A fraction whose denominator is a sum of two square roots is rationalised by multiplying numerator and denominator by the conjugate of that denominator, because (√a + √b)(√a − √b) = a − b carries no radical. Two numbers that are reciprocals of each other have product 1, so once their sum is known a sum of cubes follows from the identity x3 + y3 = (x + y)3 − 3xy(x + y), with no need to cube either number separately.

Application. Working through the given expression:

1. Rationalise x. Multiply numerator and denominator of (√7 − √5)/(√7 + √5) by the conjugate (√7 − √5). The denominator becomes 7 − 5 = 2, and the numerator becomes (√7 − √5)2 = 7 + 5 − 2√35 = 12 − 2√35. Hence x = (12 − 2√35)/2 = 6 − √35.

2. Form y. Since y = 1/x = (√7 + √5)/(√7 − √5), multiplying numerator and denominator by (√7 + √5) gives y = (12 + 2√35)/2 = 6 + √35.

3. Collect the two symmetric quantities. x + y = (6 − √35) + (6 + √35) = 12, and xy = (6 − √35)(6 + √35) = 36 − 35 = 1, which also confirms that y is the reciprocal of x.

4. Substitute into the identity. x3 + y3 = (x + y)3 − 3xy(x + y) = 123 − 3(1)(12) = 1728 − 36 = 1692.

5. Take the square root and pull out the perfect square. 1692 = 36 × 47, so √1692 = √36 × √47 = 6√47.

Cross-check. Numerically √35 ≈ 5.9161, so x ≈ 0.0839 and y ≈ 11.9161. Then x3 + y3 ≈ 0.0006 + 1691.9994 ≈ 1692, and √1692 ≈ 41.13, which agrees with 6 × √47 ≈ 6 × 6.8557 = 41.13. Because 47 is prime it holds no square factor, so 6√47 is the fully simplified surd and the value of √(x3 + y3) is 6√47.