Question

Y = cbrt24 + cbrt24+ cbrt24......till infinity . Find y

Answer 1

Solution :-

$\sf y = \sqrt[3]{24 + \sqrt[3]{24 + \sqrt[3]{24..... \infty } } }$

Cubing on both sides

$\implies \sf y^{3} = \bigg(\sqrt[3]{24 + \sqrt[3]{24 + \sqrt[3]{24}..... \infty }} \bigg)^{3}$

$\implies \sf y^{3} = 24 + \sqrt[3]{24 + \sqrt[3]{24..... \infty } }$

⇒ y³ = 24 + y

⇒ y³ - y - 24 = 0

If we divide y³ - y - 24 by y - 3 we get remainder as 0 and quotient as y² + 3y + 8 [ Refer to attachment for division. ]

So, factors of y³ - y - 3 are (y - 3)(y² + 3y + 8)

Hence, the equation can be written as

⇒ (y - 3)(y² + 3y + 8) = 0

⇒ y - 3 = 0 or y² + 3y + 8 = 0

⇒ y = 3 or y² + 3y + 8 = 0

Let's solve y² + 3y + 8 = 0

⇒ y² + 3y + 8 = 0

Discriminant = b² - 4ac = 3² - 4(1)(8) = 9 - 32 = - 23

Discriminant is less than 0, So the equation has no real roots

Therefore the value of y is 3.