Question

Let K be an integer and,

$n = \sqrt[3]{k + \sqrt{ {k}^{2} - 1} } + \sqrt[3]{k - \sqrt{ {k}^{2} - 1 } } + 1$
Then prove that:
${n}^{3} - 3 {n}^{2} \: is \: an \: integer$
PLZ ANSWER THIS QUESTION AS FAST AS POSSIBLE...FIRST CORRECT ANSWER WITH DETAILED AND CORRECT SOLUTION WILL GET BRAINLIEST ​

Answer 1

Sᴏʟᴜᴛɪᴏɴ :-

→ n = ³√[k + √(k² - 1)] + ³√[k - √(k² - 1)] + 1

→ (n - 1) = ³√[k + √(k² - 1)] + ³√[k - √(k² - 1)]

cubing both sides, we get,

→ (n - 1)³ = {³√[k + √(k² - 1)] + ³√[k - √(k² - 1)]}³

using formula Now :-

• LHS :- (a - b)³ = a³ - b³ - 3ab(a - b)
• RHS = (a + b)³ = a³ + b³ + 3ab(a + b)

→ n³ - 1 - 3n(n - 1) = √[k + √(k² - 1)] + √[k - √(k² - 1)] + 3 * √[k + √(k² - 1)] * √[k + √(k² - 1)] * (n - 1)

using (a + b)(a - b) = (a² - b²) in RHS,

→ n³ - 3n² + 3n - 1 = √[k + √(k² - 1)] + √[k - √(k² - 1)] + 3 * { k² - (k² - 1)} * (n - 1)

→ n³ - 3n² + 3n - 1 = √[k + √(k² - 1)] + √[k - √(k² - 1)] + 3 * (n - 1)

→ n³ - 3n² + 3n - 1 = √[k + √(k² - 1)] + √[k - √(k² - 1)] + 3n - 3

→ n³ - 3n² - 1 = √[k + √(k² - 1)] + √[k - √(k² - 1)] - 3

→ n³ - 3n² = √[k + √(k² - 1)] + √[k - √(k² - 1)] - 3 + 1

→ n³ - 3n² = √[k + √(k² - 1)] + √[k - √(k² - 1)] - 2

Now, since k is an integer , any value of k (whether , we put k = 0 , -1 , -2 , + 1, +2 } , we will get an integer on RHS side, as (-2) is an integer .

Hence, we can conclude that, (n³ - 3n²) is an integer.